M 9102. Introduction to Linear Polynomials
अचर :- संख्याओं या अंको से व्यक्त राशि को अचर कहते हैं। अचरों का मान अपरिवर्तनशील होता है। इन्हे हम अंक भी कहते हैं क्योंकि ये अंका के रूप में लिखे जाते हैं। जैसे 1, 2, 3, 7, 25, 100,
Constant: - The quantity represented by numbers is called constant. These are also called Numerals because these are Numbers. It value is fixed i.e. 1, 2, 3, 7, 25, 100, ............
Numeric Value or counting Numbers are called Constant.
चर: अक्षरों से व्यक्त राशि को चर कहते हैं। ये अक्षर अंग्रेजी या रोमन वर्णमाला के हो सकते हैं। इन्हे अक्षर भी कहते हैं क्योंकि ये अक्षरों के रूप में प्रयुक्त होते हैं। चरों का मान परिवर्तनशील होता है। जैसे x, y, z, a, b, c, l, m, n, α, β, γ, θ, λ etc.
Variables: The quantity represented by letters is called variables. These letters may be from Roman or English alphabets. These are also called Literals because variables are letters. Its value is not fixed i.e. x, y, z, a, b, c, l, m, n, a, β, γ, θ, λ etc. 2x, 7z, xy, yz, a
Alphabets of English and Roman are called Variables
पद :- एक बीजगणितीय एस०एन०ए० रूप में लिखा गया कुछ पद कहलाता है। एक चिन्ह वाले मान को पद कहते हैं। जैसे x, y, z, a, b, c, l, m, n, 2x, 7z, xy, yz, abc, x², yz², x²y²z² etc.
Term: - Something written algebraically in single SNA form is called a Term.
It is also called Meal. The value of single sign is known as term i.e. x, y, z, a, b, c, l, m, n, 2x, 7z, xy, yz, abc, x², yz², x²y'z etc.
Standard form of term is SNA form. S– Signs, N – Numbers and A – Alphabetically form.
Note: Term is always connected by the sign of multiplication (invisible dot) with their constituent.
ध्यान देंः पद हमेशा अपने अव्यवों के मध्य गुणा का चिन्ह (अदृश्य डॉट) छिपाए रखता है।
पदों के प्रकार: पद तीन प्रकार के होते हैं। क) अंकीय, ख) अक्षरीय तथा ग) मिश्रित या संयुक्त पद ।
Types of Terms: There are three types of terms. a) Numeric b) Alphabetic and c) Mixed
a) Numeric Term or Constant Term: - The term written in numeric form (Number) is called Numeric Term or Constant Term i.e. 2, 5, 7, 105, 9999 etc.
क) अंकीय पद :- अंकों के रूप में लिखे गए पदों को अंकीय पद कहते हैं। जैसे 2, 5, 7, 105, 9999 etc.
b) Alphabetic Term: The term written in alphabetic form is called Alphabetic Term i.e. a, a, a, a", x², m³, n" etc.
ख) अक्षरीय पद : अक्षरों के रूप में लिखे गए पदों को अक्षरीय पद कहते हैं। जैसे a, a, a, a", x², m³, n" etc.
c) Mixed Term: Alphabets The term written with the help of Number and is known as mixed term i.e. 2x, 5x², ab, xyz, xy²etc. mixed term i.e.
ग) मिश्रित या संयुक्त पद : अंको व अक्षरों से संयुक्त रूप से लिखे गए पदों को संयुक्त या मिश्रित पद कहते हैं। जैसे 2x, 5x², ab, xyz, xy²etc.
(01) निम्न में से प्रत्येक बहुपद में आए पदों को वार्णिक, आंकिक तथा मिश्र पदों में विभक्त करो।
Write the number of the terms in the form of Alphabetic, Numeric and mixed term from the each of the polynomials.
01. x³ + 8
02. x² – 30x + 9
03. x² + x
04. 5x² – 4xy –9
05. x – x²
06. 2 – x² + 7mn
07. 25x² – 30x + 9
08. 3x
09. 3x – 6
10. √3x – 1
11. 2 y²
12. 1 + x
13 x ² – x – 12
14. 2 + x ² + ab
15. x + 3
Sol.: -
01. In x³ + 8y + 5, x³ is Alphabatic term, 8y is mixed term and 5 is numeric term.
x³ + 8y + 5 में x³ वार्णिक, 8y मिक्स तथा 5 आंकिक पद है।
चरों के आधार पर पदों के प्रकार : चरों की प्रकृति के आधार पर पद दो प्रकार के होते हैं। समान पद तथा असमान पद ।
Types of the terms on the bases of variables: - There are two types of temrs on the bases of variables. Like terms and Unlike tems.
समान पद :- वे पद जिनके समान अक्षरीय गुणनखण्ड होते हैं। समान पद कहलाते हैं। जैसे abc, cab व acb समान पद हैं।
Like Terms: The terms which have same literal factors is called Like terms i.e. ab and ba are like terms. abc, cab and acb are like tems.
abc, cab and acb are not written according to SNA form if we use it then all three terms will be abc
असमान पद : वे पद जिनके अक्षरीय गुणनखण्ड समान नहीं होते असमान पद कहलाते हैं। जैसे
Unlike Terms: The terms which have not same literal factors is called Unlike terms i.e.
02. Find like terms from the following: (निम्न में से को समानपदी हैं।)
01. 7x, 14x, -13x, 5x², 7y, 7xy, -9y², -9x², -5yx
02. - ху, 7 yx², 6 x 22, 18 zx, 3x²
03. -5yx, 2xy², 6x²y², -9 x²²
04. -xy, 4yx, 8x, 2xy², 7y,
05. 11x², 100x, 11yx, 20x²y,
06. 6x y, 2xy, 3x
07. 10pq, 7p, 8q, pq, -5p², 41, 2405p, 78qp, 2 7qp, 100g, 23, 12qp, 13p²q, qp², 701p²
08. 10pq, 7p, -7qp, -100q-5p², 41, 2405p, qp², 701p²
Sol.
01. 7x, 14x, –13x are like terms.and 5x² and -9x² are like terms 2
व्यंजक :- पदों के समूह को जो गणित की आधार भूत चिन्हों के द्वारा संबंद्ध रहते हैं व्यंजक कहलाते हैं। अतः एक या उससे अधिक पदों के संयोग से व्यंजक बनता है। यह कई प्रकार का हो सकता है। जैसे एकपदीय, द्विपदीय, त्रिपदीय तथा बहुपदीय।
Expression: - Group of term connected by the sign of fundamental operations (addition, subtraction, multiplication or division) is known as Expression.
The Expression is formed by adding or subtracting one or more terms. These are of many types like monomials, binomials, trinomials and polynomials.
पदों की संख्या के आधार पर व्यंजकों के प्रकार :- पदों की संख्या के आधार पर पद चार प्रकार के होते हैं।
Types of expression on the basses of Number of terms: There are mainly four types of expressions on the bases of the Numbers of Terms.
एकपदीय :- वह व्यंजक जिसमे केवल एक पद (या एक चिन्ह) होता है। एक पदीय व्यंजक कहलाता है। जैसे x, y, z, xyz, abc, 2x, 7z, xy, yz, abc, x², yz², x²y²z² etc.
Monomials: The Expression that contains only one term (or one sign) is called monomials i.e. x, y, z, xyz, abc, 2x, 7z, xy, yz, abc, x², yz², x²yz² etc.
द्विपदीय :- वह व्यंजक जिसमे केवल दो पद (या दो चिन्ह) होते हैं। द्विपदीय व्यंजक कहलाता है। जैसे x + y, za, 2x + xyz, a + b²c, 2+x, z–2, xy + yz, abc + a²bc, x² + yz², x²+y² etc.
Binomials: The Expression that contains only two term (or two signs) is called binomials i.e. x + y, za, 2x + xyz, a + b²c, 2+x, z –2, xy + yz, abc + a²bc, x² + yz², x²+y² etc.
त्रिपदीय : वह व्यंजक जिसमे केवल तीन पद (या तीन चिन्ह) होते हैं। त्रिपदीय व्यंजक कहलाता है। जैसे x + y +2, z–a+b, 2x + xy + z, a+b+c, a+2+x, z–2–y, xy + yz–zx, abc + a²bc+bc, x² + y + z², x²+y² + z² etc.
Trinomials: - The Expression that contains only three term (or three signs) is called trinomials i.e. x + y +2, z–a+b, 2x + xy + z, a+b+c, a+2+x, z–2–y, xy + yz–zx, abc + a²bc+bc, x² + y + z², x²+y² + z² etc
बहुपद :- तीन या उससे अधिक पदों (अथवा तीन या अधिक चिन्हों) वाला व्यंजक, जिसके गुणज शून्येत्तर (शून्य न हों) तथा घातें ऋणेत्तर (ऋण न हों) बहुपद कहलाता है। जैसे x + y +2, z–a+b, 2x + xy + z, a + b² + c, a+2+x, z–2–y, xy + yz–zx, abc + a²bc+ bc, x² + y + z², x²+y² + z² etc.
Polynomials: - The Expression containing one or more terms (or with three or more sign) with non zeros coefficient and with non negative exponent is known as polynomials i.e. x + y +2, z–a+b, 2x + xy + z, a + b² + c, a+2+x, z–2–y, xy + yz–zx, abc + a²bc+ bc, x² + y + z², x²+y² + z² etc.
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बहुपद की घात :- किसी बहुपद की उच्चतम धात को बहुपद की घात कहते हैं। 4x + 2 की घात एक तथा 7m⁷ + ¾ m⁴ – 4x + 2 की घात सात है।
Power of Polynomials: - Highest power of the polynomials is called the power of the polynomials. The power of 4x+2 is one and 7m⁷ + ¾ m⁴ – 4x + 2 is seven.
(03) निम्न में से प्रत्येक बहुपद की धात ज्ञात करो।
Write the degree of each of the following polynomials:
01. 25x² – 30x + 9
02. √2x – 1
03. x¹⁰ + y³ + z⁵⁰
04. x⁵ – x³ + 3
05. x⁶ – y⁶
06. 2 – y² – y³ + y⁸
07. x⁸ – 256
08. 2 + x² + x
09. x³ + x ² * y + x y ² + y³
10. π/2 x² + x
11. x² – x – 12
12. –35x
13. 5x³ + 4x² + 7x
14. 4 – y
15. 4t – √3
बहुपद की घात :- किसी बहुपद की उच्चतम धात को बहुपद की घात कहते हैं।
Power of Polynomials: - Highest power of the polynomials is called the power of the polynomials.
Sol.
01. 25x² – 30x + 9 की घात दो है।
01. The power of 25x² – 30x + 9 is two.
03. x¹⁰ + y³ + z⁵⁰ की घात पचास है।
03. x¹⁰ + y³ + z⁵⁰ is fifty.
1. Find the degrees of the following polynomials:
(i) 2x² – 5x + 3
(ii) y³ + 2y – 1
(iii) – 9
(iv) 4z – 3
2. Write polynomials of degrees 1, 2, 3 and 4.
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गुणांक :- किसी बीजीय राशि के साथ जुडी संख्या को उस राशि का गुणांक कहते हैं। यह धनात्मक तथा ऋणात्मक हो सकता है। धन को बिन चिन्ह के तथा ऋणात्मक को ऋण चिन्ह के साथ लिखते हैं। समी० x³ – 3x² – 9x + 5 में x³ में x³ का गुणांक का गुणांक 1 तथा 3x² में, x² का गुणांक 3 है। 9x में, x का गुणांक –9 है, और अचरपद 5 है।
Coefficients: The numerical factor of the term is called the coefficient. It may be positive or negative. Positive is written without sign and negative written with negative sign. In the equation x³ – 3x² – 9x + 5 the coefficient of x³ is one and coefficient of x² is 3 and coefficient of x is –9. Hire + 5 is constant.
निम्न में से प्रत्येक में x तथा x² का गुणांक ज्ञात करो।
Write the coefficients of x and x² in each of the followings:
1. x¹⁰ + y³ + x² 2. x² – 30x + 9
3. x³ + 8 4. 5x² – 4x – 9
5. 25x² – 30x + 9 6. π/ 2 x² + x
7. x⁶ – y⁶ 8. √2 x –1
9. x⁸ – 256 10. 2 – y² – y³ + y⁸
11. x³ + x² y + x y² + y³ 12. 2 – x² + x³
13 x² – x – 12 14. 2y²
15. 2 + x² + x
गुणांक :- किसी बीजीय राशि के साथ जुडी संख्या को उस राशि का गुणांक कहते हैं।
Coefficient: The connecting digits (Numerical value) with the variable are called the coefficient of the variables.
हल :-
01. समी० x¹⁰ + y³ + x² में , x² का गुणांक 1 तथा x का गुणांक 0 है।
In the equation x¹⁰ + y³ + x² the coefficient of the x² is one and coefficient of x is 0.
02. समी० 25x² – 30x + 9 में , x² का गुणांक 25 तथा x का गुणांक –30 है।
In the equation 25x² – 30x + 9 the coefficient of the x² is 25 and coefficient of x is –30.
3. What are the coefficients of x2 and x3 in the polynomial x4 – 3x3 + 6x2 – 2x + 7?
4. What is the coefficient of z in the polynomial 4z3 + 5z2 – 11?
5. What is the constant term of the polynomial 9x3 + 5x2 – 8x –10?
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समीकरण (व्यंजक) का मानक रूप :- किसी समीकरण को उसकी घातों के घटते क्रम में रखने पर प्राप्त रूप को मानक रूप कहते हैं। इस सूत्र में भी बहुपद मानक रूप में लिख हुआ है
देखिए ax² + bx + c या ax² + bx¹ + cx⁰ बहुपद या समीकरण x³ – 3x² – 9x – 5 मानक रूप में लिखा है।
Standard form of Equation (Polynomial): - Writing of any equation in the order of decreasing powers is known as standard form of equation. It can be written in the form of ax² + bx + c or ax² + bx¹ + cx⁰. The polynomial x³ – 3x² – 9x – 5 is in the standard form.
To wtite the standard form expression or polynomial in standard form write the ter m in decreasing order of power.
व्यंजक को मानक रूप में लिखने के व्यंजक के पदों को घातों के घटते क्रम में लिखते हैं।
(05) निम्न में से प्रत्येक व्यंजक को मानक रूप में लिखो। Write the Following Expression in standard form.
1. x¹⁰ + y³ + x²
2. –30x⁵ + x² – 30x + 9
3. x³ + 8
4. 5x² – 4x – 9
5. 25x² – 30x + 9
6. π/ 2 x² + x
7. x⁶ – y⁶
8. √2 x –1
9. x⁸ – 256
10. 2 – y² – y³ + y⁸
11. x³ + x² y + x y² + y³
12. 2 – x² + x³
13 x² – x – 12
14. 2y²
15. 2 + x² + x
Sol 1. x ¹⁰ + y³ + x²
Sol. 3. –30x⁵ + x² – 30x + 9
बहुपद :- वह व्यंजक जिसके गुणज शून्येत्तर (शून्य न हों) तथा घांतें ऋणेत्तर (ऋण न हों) बहुपद कहलाता है। जैसे x + y + 2 z – a + b, 2x + xy+z, a+b²+c, a+2+x, z-2-y, xy + yz-zx, abc + a²bc+bc, x² + y + z², x²+y² + z² etc. किसी बहुपद का सर्वसामान्य रूप है। a_{0} + a_{1} * x ^ 1 + a_{2} * x ^ 2 +..........+a n x^ n
Polynomials: - The Expression containing one or more terms with non zeros coefficient with non negative exponent is known as polynomials i.e. x + y + 2 zab, 2x + xy + z, a + b² + c a + 22 +x, z-2-y, xy + yzzx, abc + a²bc+ bc, x ^ 2 + y + z ^ 2, x ^ 2 + y ^ 2 + z ^ 2 * et The general form of a polynomial in one variable is a_{0} + a_{1} * x ^ 1 + a_{2} * x ^ 2 +.........+a n x^ n
(06) State which of the following are monomials, binomials, trinomials and polynomials:
निम्न में से एक पदीय, द्विपदीय त्रिपदीयए बहुपदीय व्यंजकों को छांटोः
निम्न में से एक पदीय, द्विपदीय त्रिपदीयए बहुपदीय व्यंजकों को छांटोः
1. 4x – 3y 2. 7xyz 3. x² + 2x + 7
4. – z + 5 5. 7 6. a+b
7. x + 3 8. 41+5m 9. 4p²q –4q²p + r + 5
10. a + 4 11. 2y – 5, 12. 5 – 3xy
13. 3x², 14. 22 – 4y² 15. 4xy +7,
16. 4x² 17. x, 18. 3xy – y² + x² – y + x
19. x – 4, 20. –7z 21. 2 + 1
22. 5xy. 23. 3x – 2, 24. 10y – 9
25. 17, 26. 82a + b + c
27. 2x + 3y + 7z + 7. 28. 2x+3y –5
29. x + y + z + 2
30. x²y – xy² + y² – y² + x²
Solution
1. 4x –3y
It has two terms so that it is Binomial
2. 7xyz
It has one terms so that it is Monomial
3. x² + 2x + 7
It has three terms so that it is Trinomial
घातों की संख्या के आधार पर व्यंजकों के प्रकार :– घातों की संख्या के आधार पर व्यंजक चार प्रकार के होते हैं।
Types of expression on the bases of Power (Degree): There are main four types of expressions on the bases of the Numbers of the base of Power of Expression.
रैखिक बहुपद :- वह बहुपद जिनकी सबसे बड़ी घात एक हो, उस बहुपद को रैखिक बहुपद कहते हैं। यदि इसमें एक चर है तो एक चर वाला रैखिक बहुपद (ax + b, 4x + 2, m + 5) तथा दो चर है तो इसे दो चर वाला रैखिक बहुपद (ax + b, 4x + 2, m + 5) कहते हैं। जैसे
Linear Polynomials: Polynomial with power one is called the linear polynomials. If it has one variable then it is called the linear polynomials of one variable (ax + b, 4x + 2, m + 5) and if it has two variables then it is called the linear polynomials of two variables and SO on.
(ax + by + c, 2x + 3y + 4), 4x + 2 , √5x + 5 , x –¾ , ¾m + 1
रैखिक बहुपद : घात 1 वाले बहुपद को रैखिक बहुपद कहते हैं। जैसे
Linear Polynomial:- A polynomial of degree 1 is called a linear polynomial. As
द्विघात बहुपदः वह बहुपद जिनकी सबसे बड़ी
घात दो हो, उसे बहुपद को द्विघात बहुपद कहते हैं। जैसे x² – 4x – 12, x² + 2x + 3, x² + √5x + 5
Quadratic Polynomials: The polynomials which have highest power two is called the quadratic polynomials. x ² – 4x – 12, x² + 2x + 3, x² + √5x + 5
द्विघात बहुपद का लिखना :- कोई भी द्विधात बहुपद को ax² + bx + c के रूप में लिखा जाता है। जहाँ, a को x² का गुणांक, b का x का गुणांक तथा c को अचरपद कहते हैं।
नोट:- ध्यान रहें कि a, b, c संख्यात्मक होते है। चर को ध्यान में रखकर हम बहुपद को निम्न प्रकार से लिख सकते हैं :
चर X वाले बहुपद P(x) = x² + 2x – 3
चर m वाले बहुपद P(m) = m ³ + 2m – 3
चर p वाले बहुपद P(p) = p² + 2p – 3
चर 2 वाले बहुपद P(z) = z² + 2z – 3
चर वाले बहुपद P(y) = y ² + 2y – 3
Writing of Quadratic Polynomial: Any polynomials can be written in the form of ax² + bx + c where 'a' is the coefficient of x² and 'b' is the coefficient of x and 'c' is constant term.
Polinomial of variable x is P(x) = x² + 2x – 3
Polinomial of variable m is P(m) = m ² + 2m – 3
Polinomial of variable p is P(p) = p² + 2p – 3
Polinomial of variable z is P(x) = z² + 2z – 3
Polinomial of variable y is P(y) = y² + 2y – 3
Note: a, b and c are written in numerical value.
त्रिघात बहुपद :- वह बहुपद जिनकी सबसे बड़ी घात तीन हो उस बहुपद को त्रिघात बहुपद कहते हैं। x³ – 3x² – 9x – 5, x³ + 12x² + 32x + 20, x³ – 2x² – x + 2
Cubical Polynomials: The polynomials which have highest power three is called the cubical polynomials.
x³ – 3x² – 9x – 5, x³ + 12x² + 32x + 20, x³ – 2x² – x + 2
चतुर्थघातीय बहुपद :- घात चार वाले बहुपद को चतुर्थघातीय बहुपद कहते हैं।
2x⁴ – x³ – 2x² – x + 2; 2x⁴ + x³ + 12x² + 32x + 20; 5x⁴ + x³ – 3x² – 9x – 5
Biquadrate Polynomials: - The polynomials which have highest power four is called the biquadrate polynomials.
2x⁴ – x³ – 2x² – x + 2; 2x⁴ + x³ + 12x² + 32x + 20; 5x⁴ + x³ – 3x² – 9x – 5
आंकिक बहुपद :- जब किसी बहुपद में केवल एक ही पद वह भी अंक हो तो इस प्रकार के बहुपद को हम आंकिक बहुपद कहते हैं। 2, 5, 7, आदि आंकिक बहुपद के उदाहरण हैं।
Constant polynomials: - If there is only one constant in the polynomial then it is called constant polynomial. 2, 5, 7, etc. are examples of constant polynomials.
शून्य बहुपद :- यदि किसी बहुपद में उसके गुणांक शून्य हों तो उस बहुपद को शून्य बहुपद कहते हैं। (a0 = a1 = a2 = 0)
Zero polynomial: - If all the constant a0 = a1 = a2 = 0) of polynomial is,
Zero then constant polynomial 0 is called the zero polynomial.
(07) निम्न में से प्रत्येक बहुपद को रेखिक, द्विघातीय बहुपद, त्रिघात बहुपद, आंकिक तथा शून्य बहूपदों में विभक्त करो।
Classify the following as linear, quadratic, cubic, constant or zero polynomials.
1. x³ + 8 2. x² – 30x + 9
3. x² + x. 4. 5x² – 4x – 9
5. x – x³ 6. 2 – x² + x
7. 25x² – 30x + 9 8. 3x
9. 1 + x. 10. √2 x – 1
11. 3x –6 12. x + 3
13. x² – x – 12 14. 2 y²
15. 2 + x² + x 16. –7y
17. 7 18. 0
19. –11 20. 0 m²
21. y + y² + 4 22. 3t
23. 7² 24. 7³
Sol. Do yourself by looking the highest power if power is one then it is linear polynomials, if power is two then it is quadratic polynomials and if power is three then it is cubic polynomials. If there is only number than it is constang polynomial and it there is 0 then is zero polynomial.
आप स्वयं करें क्योंकि घात 1 वाले बहुपद को रैखिक बहुपद कहते हैं। घात 2 वाले बहुपद को द्विघात बहुपद कहते हैं। घात 3 वाले बहुपद को त्रिघात बहुपद कहते हैं। घात 0 वाले बहुपद को आंकिक बहुपद कहते हैं। शून्य मान वाले बहुपद को शून्य बहुपद कहते हैं।
(08) निम्न में से कौन एकचर में बहुपद है और कौन नहीं।
Write which of the following are polynomials of one variable and which are not.
01. z² – 2z – 8
02. x² – 30x + 9
03. x¹⁰ + y³ + z³⁰
04. 5x² – 4x – 9
05. x – x³
06. π/2 x + x
07. 25x² – 30x + 9
08. 3x
09. –7y
10. 2 – y² – y³ + y⁸
11. x + 3
12. x³ + x² y + x y² + y³
13 x² – x –12
14. 2y²
15. 2 + x² + x
16. 4x² – 3x + 7
17. y² + √2
18. x¹⁰ + y³ + t⁵⁰
हलः- वह बहुपद जिसमें केवल एक ही प्रकार के चर हो एक चर में बहुपद कहलाता है।
01. z² – 2z – 8 एक चर में बहुपद है क्योकि इसमें केवल एक ही चर 'z' है।
In this polynomial there is, only one variable 'z' so it is polynomial of one variable 'z'.
The degree of a non-zero constant polynomial is zero.
बहुपद का शून्य या शून्यक या गुणनखंड या हल : वे मान जो समीकरण में प्रतिस्थापित करने पर उसे शून्य में बदल देते हैं। बहुपद के शून्य, शून्यक, गुणनखंड या हल कहलाते हैं। घ्यान रहे कि किसी बहुपद की घात के बराबर उसके हल या शून्य हो सकते हैं।
Zeros or root or solution of polynomials: A value of the variable which makes the polynomial zero is called the zero, root or solution of the polynomials.
Example 1: Raju went to a shop where there were sealed boxes of different colours on sale. The shop owner told him that the red boxes have 4 pens each and the blue boxes have 5 pencils each. Now, if Raju bought x red boxes and y blue boxes, how can he quickly figure out the total quantity of pens and pencils? Also, if he got 3 extra pens free, how many pens and pencils did he get altogether?
Sol.
Let:
Number of red boxes = x
Number of blue boxes = y
Each red box has 4 pens, so total pens = 4x
Each blue box has 5 pencils, so total pencils = 5y
Therefore, total number of pens and pencils is: = 4z + 5y
If Raju gets 3 extra pens free, then total becomes: 4x + 5y + 3
Terms. Variables and Coefficients
Terms: 4z, 5y, 3
Variables: z, y
Coefficients: 4, 5
Content: 3
Example 2: A rectangular garden of length l metres and width w metres has to be fenced and decorated. A wire fence is to be laid along the length costing `100 per metre and a wooden fence is to be built along the width costing `80 per metre. Special seeds have to be sown throughout the garden which will cost `50 per square metre.
Sol.
Length of garden = I metres
Width of garden = w metres
Cost of wire fence along length 100l
Cost of wooden fence along width = 80w
Cost of seeds
Area of rectangle: l × w
Seed cost: 50/w
Total cost = 100l + 80w + 50lw
Think and Reflect
1. Terms: 100l, 80w, 50lw
Variables: l, w
Coefficients: 100, 80, 50
2. In Example 1, variables are separate (x, y). In Example 2, one term contains multiplication of variables (lw).
Think and Reflect
1. Can you identify the terms, variables and coefficients of this algebraic expression?
2. How is it different from the algebraic expression in Example 1?
Example 3: A wire of length 20 cm is bent in different ways to form rectangles. For example, we can have a rectangle with length 7 cm and width 3 cm. We can also have one of length 5.5 cm and width 4.5 cm.
(Think of a few more ways of forming such rectangles.) Can you write an expression for the area of such rectangles?
Sol.
Total wire length = 20 cm
Perimeter of rectangle: 2(l +w) = 20
So,
l + w = 10
If length l = x,
then width: = 10 –x
Area of rectangle = x(10 –x)
= 10x – x²
So, the algebraic expression for area is: 10x –x²
Think and Reflect
1. Terms: 10x, –x²
Variable: x
Coefficients: 10, –1
2. Similarity:
Both examples use variables and algebraic expressions.
Difference:
Example 1 has only first power variables. Example 3 contains z², so it is a quadratic expression.
Think and Reflect
1. Can you identify the terms, variables and coefficients of this algebraic expression?
2. Can you point out any similarity or difference between the algebraic expressions obtained in Examples 1 and 3?
Example 4: The perimeter of a square of side x is 4x, which is a linear polynomial in the variable x.
Sol.
Perimeter of square p : 4x
Think and Reflect
Find the perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm and 3 cm. What will happen to the perimeters if the sides increase by 0.5 cm?
Sol.
Find perimeters
Side (cm) Perimeter 4x (cm)
1 4
1.5 6
2 8
2.5 10
3 12
Observation
Whenever the side increases by 0.5 cm, the perimeter increases by:
4 × 0.5 = 2 cm
Example 5: A chess club charges a joining fee of `200 plus `50 for every match played. The following table shows the amount a player will have to pay as the number of matches varies.
Sol.
Joining fee = ₹200
Cost per match = ₹50
If number of matches = x
Then total amount paid = 200 + 50x
Example 6: The sum of two numbers is 64. One of the numbers is 10 more than the other. What are the two numbers?
Sol.
Let the smaller number be = x
Then the larger number = x + 10
The sum of both numbers = 64
A.T.Q.
x + ( x + 10) = 64
x + x + 10 = 64
2x + 10 = 64
2x = 64 – 10
2x = 54
x = 27
So the larger number:
= x + 10
= 27 + 10
= 37
Ans. The two numbers are 27 and 37
Think and Reflect
We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable in the given expression. Consider Example 3, where the wire is bent to form a rectangle. Here, the area of the rectangle, 10x – x², is a function of x. Can you interpret this as an input-output process? What value does the expression take when x = 6 cm?
Sol.
Final Think and Reflect
In Example 3, area = 10x –x² ........(1)
This works like an input-output process:
Input value of x
Output = area of rectangle
When x = 6
Now, Substitute x = 6 in equation (1)
10x –x²
= 10(6) – 6²
= 60 – 36
= 24
Ans. When x = 6 cm, the area is 24 cm
Exercise Set 2.2
Q. 1. Find the value of the linear polynomial 5x – 3 if:
(i) x = 0 (ii) x = –1 (iii) x = 2
Sol.
1. Find the value of the linear polynomial 5x – 3
(1) When x = 0
= 5x – 3
= 5(0) – 3
= 0 – 3
= – 3
Answer. –3
(ii) When x = –1
= 5x – 3
= 5(–1)–3
= –5 – 3
= – 8
Answer. –8
(iii) When = 2
= 5x – 3
= 5(2) –3
= 10 –3
= 7
Answer: 7
Q. 2. Find the value of the quadratic polynomial 7s² – 4s + 6 if:
(i) s = 0 (ii) s = –3 (iii) s = 4
Sol.
(i) When s = 0
Then
7s² – 4s + 6
= 7(0)² –4(0) +6
= 7(0) – 0 + 6
= 0 – 6
Answer: 6
(ii) When s = –3
Then
7s² – 4s + 6
= 7(–3)² –4(–3) +6
= 7(9) + 12 + 6
= 63 + 12 + 6
= 81
Answer: 81
(iii) When s = 4
Then
7s² – 4s + 6
= 7(4)² –4(4) +6
= 7(16) – 16 + 6
= 112 – 10
= 102
Answer: 102
Q. 3. The present age of Salil’s mother is three times Salil’s present age. After 5 years, their ages will add up to 70 years. Find their present
ages.
Sol.
3. Let Salil's present age be x years.
Then mother's age = 3x
After 5 years:
Salil's age = x + 5
Mother's age = 3x + 5
According to the question (A.T.Q.):
(x + 5) + ( 3x + 5) = 70
=> x + 5 + 3x + 5 = 70
=> 4x + 10 = 70
=> 4x = 70 – 10
=> 4x = 60
x = 60/4
x = 15
So,
Salil's present age 15 years
Mother's present age = 3 × 15
= 45
Ans.
Salil 15 years
Mother 45 years
Q. 4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
Sol.
Difference between two integers is 63 and
ratio is 2:5
Let the integers be 2x and 5x, then
Difference:
5x – 2x = 63
3x = 63
x = 63/3
x = 21
Then First integer is
2x
= 2(21)
= 42
Second integer is
5x
= 5 × (21)
= 105
Answer. The integers are 42 and 105
Q. 5. Ruby has 3 times as many two-rupee coins as she has five rupee-coins. If she has a total `88, how many coins does she have of each type?
Sol.
Value of two-rupee coins = 2x
Then number of two-rupee coins = 3
Value of two-rupee coins = 3(2x)
= 6x
Value of five-rupee coins = 5x
Total money:
5x + 6x = 88
11x = 88
x = 88/11
x = 8
So,
Five-rupee coins = 8
Two-rupee coins = 3x
= 3 (8)
=24
Answer
Five-rupee coins = 8
Two-rupee coins = 24
Q. 6. A farmer cuts a 300 feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How
long are the two pieces?
Sol.
Let the shorter piece be x feet.
Longer piece = 4x
A.T.Q.
Total length
=> x + 4x = 300
=> 5x = 300
x = 300/5
x = 60
So,
Shorter piece = 60 feet
Longer piece: = 4x
= 4 × 60
= 240
Answer:
Shorter piece = 60 feet
Longer piece = 240 feetp
Q. 7. If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle?
Sol.
Let width = x cm
Length = 2x + 3 cm
Perimeter of rectangle:
2(l + b) = 24
=> 2 [(2x + 3) + x] = 24
=> 2 [2x + 3 + x] = 24
=> 2 [3x + 3] = 24
=> 6x + 6 = 24
=> 6x + 6 = 24 – 6
6 x = 18
x = 18/6
x = 3
Width = 3
Length = 2x + 3
= 2 × 3 +3
= 6 + 3
= 9
w = 3
l = 9
Example 7: Bela has ₹100 for pocket money. She spends ₹5 every day. After how many days will she be left with ₹40?
Bela has ₹100 as pocket money. She spends ₹5 every day.
We need to find after how many days she will have ₹40 left.
Let the number of days be x
Money left after x days = 100 – 5x
According to the question:
100 – 5x = 40
Subtract 100 from both sides:
-5x = -60
x = –60/–5
x = 12
Answer:
Bela will be left with ₹40 after 12 days.
Example 8: An auto-rikshaw fare starts at 25 and remains the same for the initial 2 km. Then it increases by ₹15 per km. What will be the fare for a travel of 10 km?
Initial fare for first 2km = overline r 25
After that, fare increases by ₹15 per km.
Let total distance travelled be x km.
A.T.Q.
The fare equation becomes:
F = 25 + 15(x – 2)
For
x = 10km
F = 25 + 15(x – 2)
Substitute x = 10
F = 25 + 15(10 – 2)
F = 25 + 15(8)
F = 25 + 120
F = 145
Ans. The fare for 10 km is ₹145.
2nd Method
Auto-rikshaw fare:
Initial or fixed 2 km fare = ₹25
After 2 km, fare increases by ₹15 per km
Distance travelled = 10 km
Extra distance after first 2 km:
10 – 2 = 8 km
Extra fare: 8 × 15 = 120
Total fare:
25 + 120 = 145
Answer:
The fare for 10 km travel will be ₹145.
Example 9: The cost of a journey is given by the linear function C(d) = 100+ 60d, where C indicates total cost in rupees and d the distance travelled in km. Let us make a table of values for d varying from 0 to 10 km and show how the cost increases for every km.
Sol.
Given:
C(d) = 100 + 60d
where:
C(d) = total cost in rupees
d = distance travelled in km
We increase this table for d = 0 to 10.
| Distance (km) | Cost |
|---|---|
| 0 | ₹100 |
| 1 | ₹160 |
| 2 | ₹220 |
| 3 | ₹280 |
| 4 | ₹340 |
| 5 | ₹400 |
| 6 | ₹460 |
| 7 | ₹520 |
| 8 | ₹580 |
| 9 | ₹640 |
| 10 | ₹700 |
Observation:
The cost increases by ₹60 for every extra kilometre.
Example 10: The height of water in a cylindrical tank is 3 m at the start of summer. The height h m at the end of t months is given by the linear function h(t) = 3 – 0.5 t.
The height of water after months is
h(t) = 3 – 0.5 t
Here,
h (t) = height of water (in metres)
t = time in months
Understanding the function
At the start of summer:
h(0) = 3 – 0.5 (0) = 3
So, the initial height of water is 3 m.
Every month, the water level decreases by 0.5 m.
Find the height after different months
| Months | Height |
|---|---|
| 0 | 3 – 0.5 (0) = 3 m |
| 1 | 3 – 0.5 (1) = 2.5 m |
| 2 | 3 – 0.5 (2) = 2 m |
| 3 | 3 – 0.5 (3) = 1.5 m |
| 4 | 3 – 0.5 (4) = 1 m |
| 5 | 3 – 0.5 (5) = 0.5 m |
| 6 | 3 – 0.5 (6) = 0 m |
Conclusion
The water level decreases uniformly every month, and the tank becomes empty after 6 months.
EXERCISE SET 2.3
Q. 1. A student has ₹500 in her savings bank account. She gets ₹150 every month as pocket money. How much money will she have at the end of every month from the second month onwards? Find a linear expression to represent the amount she will have in the nth month.
Sol.
A student already has ₹500 in her bank account.
Every month she gets 150 as pocket money.
Amount after each month
After 1 month
= 500 + 150
= 650
After 2 months
= 500 + 2(150)
= 500 + 300
= 800
After 3 months
= 500 + 3(150)
= 500 + 450
= 950
After 4 months
= 500 + 4(150)
= 500 + 600
= 1100
So the pattern is:
= 650, 800, 950, 1100,...
Linear expression for the nth month
If n is the number of months, then = 500 +150n
This gives the amount in the nth month.
Q. 2. A rally starts with 120 members. Each hour, 9 members drop out of the group. How many members will remain after 1, 2, 3, ... hours? Find a linear expression to represent the number of members at the end of the nth hour.
Sol.
A rally starts with 120 members.
Every hour, 9 members leave the group.
Members remaining
After 1 hour
= 120 – 9
= 111
After 2 hours
= 120 – 2(9)
= 120 –18
= 102
After 3 hours
= 120 – 3(9)
= 120 –27
= 93
After 4 hours
= 120 – 4(9)
= 120 –36
= 84
Pattern:
111, 102, 93, 84,...
Linear expression for the nth hour
= 120 – 9(n)
This gives the number of members after n hours.
Q. 3. Suppose the length of a rectangle is 13 cm. Find the area if the breadth is (i) 12 cm, (ii) 10 cm, (iii) 8 cm. Find the linear pattern representing the area of the rectangle.
Sol.
Length of rectangle = 13 cm
Area of rectangle
Area = Length × Breadth
(i) Breadth = 12 cm
Area = Length × Breadth
= 13 x 12
= 156 cm²
(ii) Breadth 10 cm
Area = Length × Breadth
= 13 x 10
= 139 cm²
(iii) Breadth 8 cm
= 13 x 8
= 104 cm²
Linear pattern
If breadth = b
then area = 13 b
Q. 4. Suppose the length of a rectangular box is 7 cm and breadth is 11 cm. Find the volume if the height is (i) 5 cm, (ii) 9 cm, (iii) 13 cm. Find the linear pattern representing the volume of the the rectangular box.
Sol.
Length = 7 cm
Breadth = 11 cm
Volume formula:
Volume = Length x Breadth x Height
= 7x11xh
= 77h
(i) Height = 5 cm
V = L × B × H
= 7 × 11 × 5
= 77 × 5
= 385 cm³
(ii) Height = 9 cm
V = L × B × H
= 7 × 11 × 9
= 77 × 9
= 693 cm³
(iii) Height 13 cm
V = L × B × H
= 7 × 11 × 13
= 77 × 13
= 1001 cm³
Linear pattern
If height = h.
Then volume = 77 H
Q. 5. Sarita is reading a book of 500 pages. She reads 20 pages every day. How many pages will be left after 15 days? Express this as a linear pattern.
Sol.
Total pages in the book = 500
Pages read every day = 20
Pages left after 15 days
Pages read in 15 days = 20 × 15
= 300
Pages left after 15 days
= 500 –300
= 200
Ans. 200 pages are left.
Linear pattern
If Sarita reads for n days,
Pages read = 20n
Pages left = 500 – 20n
This is the required linear expression.
LINEAR GROWTH AND LINEAR DECAY
EXERCISE SET 2.4
Q. 1. Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.
(i) Find the height after 7 months.
(ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.
(iii) Find an expression that relates h and t, and explain why it represents linear growth.
Sol.
(i) Height after 7 months
Initial height = 1.75 feet
Growth every month = 0.5 feet
After 7 months:
h = 1.75 + 0.5 × t
h = 1.75 + (0.5 x 7)
h = 1.75 + 3.5
= 5.25 feet
Ans. The height after 7 months is 5.25 feet.
(ii) Table of Values
Months (t) Height h (feet)
0 1.75
1 2.25
2 2.75
3 3.25
4 3.75
5 4.25
6 4.75
7 5.25
8 5.75
9 6.25
10 6.75
(iii) Expression relating h and t
h = 1.75 + 0.5 × t
It represents linear growth because the height increases by the same amount (0.5 feet) every month.
Q. 2. A mobile phone is bought for ₹10,000. Its value decreases by *800 every year.
(i) Find the value of the phone after 3 years.
(ii) Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time.
(iii) Find an expression that relates v and t, and explain why it represents linear decay.
Sol.
2. Mobile Phone Depreciation
A mobile phone costs = ₹10,000 and
its value decreases by = ₹800 every year.
(i) Value after 3 years
V = 10000 – (800 × t)
V = 10000 – (800 × 3)
= 10000 – 2400
= 7600l
Answer: The value after 3 years is ₹7,600.
(ii) Table of Values
Years (1) Value v (2)
0 10000
1 9200
2 8400
3 7600
4 6800
5 6000
6 5200
7 4400
8 3600
(iii) Expression relating v and t
V = 10000 – 800 × t
It represents linear decay because the value decreases by the same amount (₹800) every year.
Q. 3. The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.
(i) Find the population of the village after 6 years.
(ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.
(iiii) Find an expression that relates P and t, and explain why it represents linear growth.
Sol.
3. Population of a Village
Initial population = 750 Every year, 50 people move into the village.
(i) Population after 6 years P750+(50 x 6)
P=750+50t P750+300=1050
Answer: The population after 6 years is 1050.
(ii) Table of Values
Years (t) Population P
0 750
1 800
2 850
3 900
4 950
5 1000
6 1050
7 1100
8 1150
9 1200
10 1250
(iii) Expression relating P and t
P = 750 + 50 × t
It represents linear growth because the population increases by the same number (50 people) every year.
Q. 4. A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge.
(i) Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.
(ii) After how many days will the balance run out?
(iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x), reduces with time.
Sol.
Initial balance = ₹600
Balance decreases by = ₹15 every day.
(i) Equation for remaining balance
b(x) = 600 – 15 × x
It represents linear decay because the balance decreases by the same amount (15) every day.
(ii) After how many days will the balance run out?
Balance becomes zero when:
=> 600 – 15x = 0
15x = 600
x = 600/15
x = 40
Answer: The balance will run out after 40 days.
(iii) Table of Values
Days (x) Balance b(x) (₹)
1 585
2 570
3 555
4 540
5 525
6 510
7 495
8 480
9 465
10 450
Example 11: A telecom company charges a fixed monthly fee and an additional cost per GB of the internet data used. A student observes that when she used 10 GB, her bill was ₹350. When she used 20 GB, her bill was ₹550. If the monthly bill y depends on the amount of data used, x (in GB), according to the relation y = ax + b, find the values of a and b.
Sol.
Note that
x = number of GB of the internet data used and
y = total monthly cost in Rupees.
To find the linear relationship y = ax + b,
we note that when
x = 10,
y = 350
Also, when
x = 20,
y = 550
We substitute these in
y = ax + b to arrive at the following equations.
350 = 10a + b ...........(1)
550 = 20a + b ...........(2)
Subtract equation (1) from equation (2)
(550 – 350) = (20a + b) – (10a + b)
200 = 20a + b – 10a – b
200 = 10a
10a = 200
a = 200/10
a = 20
Substitute a = 20 into equation (1)
350 = 10a + b ...........(1)
350 = 10(20) + b
350 = 200 + b
b = 350 –200
b = 150
Ans = 20 and b = 150
Example 12: Let us plot the points (–1, –3), (0, 0), (1, 3), (3, 9), (4, 12) in the coordinate plane on a graph paper as shown in Figure Join the points (–1, –3) and (4, 12) using a ruler. Doing so, observe that all five points lie on a straight line. Can you guess the equation of this line by looking at the relationship between the x and y coordinates of each point?
Sol.
Example 13: Let us plot the points (–3, 6), (–2, 4), (0, 0), (1, 2), (2, –4), (3, –6) in the coordinate plane on a graph paper as shown in Fig. 2.7. Join the points (–3, 6) and (3, –6) using a ruler. Doing so, observe that all five points lie on a straight line. Can you guess the equation of this line by looking at the relationship between the x and y coordinates of each point?
Example 14: Draw the graphs of y = 1/2 x, y = x, y = 2x by selecting suitable points on these lines.
(Hint: In order to graph y = 1/2 x, we could take the points (0, 0) and (4, 2). Can you verify that these lie on the line?)
Example 15: Now let us draw the graphs of y = – 1/3 x y = – x y = – 3x by selecting suitable points on these lines. Fig. 2.10 shows the graphs of these linear equations without any points labelled on them.
Example 16: Let us now draw the graphs of y = 2x – 1, y = 2x + 1 , y = 2x + 5 first individually (as shown in Fig. 2.12) and then on the same axis (as shown in Figure).
EXERCISE SET 2.5
Q. 1. A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b find the values of a and b.
Sol.
The relation is:
y = ax + b
Where:
x = number of modules accessed
y = monthly bill
Given:
When x = 10, y = 400
When x = 14, y = 500
So,
400 = 10a + b .........(1)
500 = 14a + b .........(2)
y = ax + b
Subtract the first equation from the second:
500 = 14a + b .........(3)
± 400 = ± 10a ± b .........(1)
100 = 4a
4a = 100
a = 100/4
a = 25
400 = 10a + b .........(1)
Now substitute a = 25 into equation (1)
400 = 10a + b
400 = 10(25) + b
400 = 250 + b
250 + b = 400
b = 400 – 250
b = 150
Ans. a = 25, b = 150
Therefore, the relation is: y = 25x + 150
Q. 2. A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹1100. If the monthly bill y depends on the hours of the use of the badminton court, x, according to the relation y = ax + b , find the values of a and b.
Sol.
The relation is:
y = ax + b
Where:
2 hours of badminton court use y monthly bill
Given:
When x = 10, y = 800
When x = 15, y = 1100
So,
800 = 10a + b .........(1)
1100 = 15a + b .........(2)
y = ax + b
Subtract the first equation from the second:
1100 = 15a + b .........(2)
± 800 = ±10a ± b .........(1)
300 = 5a
5a = 300
a = 300/5
a = 60
Now substitute a = 60 into equation (1)
800 = 10a + b
800 = 10(60) + b
800 = 600 + b
600 + b = 800
b = 800 – 600
b = 200
Answer:
a = 60, b = 200
Therefore, the relation is: y = 60x + 200
Q. 3. Consider the relationship between temperature measured in degrees Celsius (°C ) and degrees Fahrenheit (°F), which is given by °C = a °F + b . Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.
(Hint: When °C = 0°, F = 32 and when °C = 100 ;° F = 212 Use this information to find a and b, and thus, the linear relationship between °C and °F.)
Sol.
Relationship Between Celsius and Fahrenheit
Given relation:
⁰C = aF + b
Where:
C temperature in Celsius
F = temperature in Fahrenheit
Given:
1. Ice melts at = 0⁰C, F = 32
So,
0⁰C = 32a + b .........(1)
2. Water boils at = 100 ⁰C, F = 212
So,
100⁰C = 212a + b .........(2)
⁰C = aF + b
Subtract the first equation from the second:
100⁰C = 212a + b .........(2)
± 0⁰C = ± 32a ± b .........(1).
100 – 0 = 212a – 32a
100 = 180a
180a = 100
a = 100/180
a = 5/9
Now substitute a = 5/9 into equation (1)
0 = 32a + b
0 = 32×9/5 + b
b = –160/9
Answer:
a = 5/9
b = –160/9
Therefore, the linear relation is:
C = (5/9) F – 160/9
This can also be written as:
C = 5/9(F – 32)
Example 5: A chess club charges a joining fee of `200 plus `50 for every match played. The following table shows the amount a player will have to pay as the number of matches varies.
EXERCISE SET 2.6
1. Draw the graphs of the following case, reflect on the role of 'a' and 'b'. ets of lines.
(i) y = 4x, y = 2x, y = x
(ii) y = – 6x, y = –3x, y = – x
(iii) y = 5x, y = – 5x
(iv) y = 3x – 1, y = 3x + 1, y = 3x
(v) y = – 2x – 3, y = 2x + 3, y = – 2x
✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️
Sol.
(i) Graphs of
y = 4x ..........(1)
y = 2x ..........(2)
y = x ..........(3)
From equation (1)
y = 4x
(1) If x = 1
Then y = 4x
y = 4 × 1
y = 4
(2) If x = 2
Then y = 4x
y = 4 × 2
y = 8
(3) If x = –1
Then y = 4x
y = 4 × –1
y = –4
Table
Points A B C
x 1 2 –1
y 4 8 –4
Coordinates (1,4) (2,8) ( –1,–4)
From equation (2)
y = 2x
(1) If x = 1
Then y = 2x
y = 2 × 1
y = 2
(2) If x = 2
Then y = 2x
y = 2× 2
y = 4
(3) If x = –1
Then y = 2x
y = 2 × –1
y = –2
Table
Points L M N
x 1 2 –1
y 2 4 –2
Coordinates (1,2) (2,4) ( –1,–2)
From equation (3)
y = x
(1) If x = 1
Then y = x
y = 1
(2) If x = 2
Then y = x
y = 2
(3) If x = –1
Then y = –x
y = –1
Table
Points X Y Z
x 1 2 –1
y 1 2 –1
Coordinates (1,1) (2,2) ( –1,–1)
Reflection
All lines pass through the origin.
Greater value of a means a steeper line.
Since a is positive, all lines rise from left to right.
✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️
(ii) Graphs of
y = –6x ..........(1)
y = –3x ..........(2)
y = x ..........(3)
From equation (1)
y = –6x
(1) If x = 1
Then y = –6x
y = –6 × (1)
y = –6
(2) If x = 2
Then y = –6x
y = –6 × (2)
y = –12
(3) If x = –1
Then y = –6x
y = –6 × (–1)
y = 6
(4) If x = –2
Then y = –6x
y = –6 × (–2)
y = 12
Table
Points A B C
x 1 2 –1
y –6 –12 6
Coordinates (1,–6) (2,–12) ( –1,6)
From equation (2)
y = –3x
(1) If x = 1
Then y = –3x
y = –3 × (1)
y = –3
(2) If x = 2
Then y = –3x
y = –3 × (2)
y = –6
(3) If x = –1
Then y = –3x
y = –3 × (–1)
y = 3
Table
Points L M N
x 1 2 –1
y –3 –6 3
Coordinates (1,–3) (2,–6) ( –1,3)
From equation (3)
y = –x
(1) If x = 1
Then y = – x
y = – 1
(2) If x = 2
Then y = – x
y = – 2
(3) If x = –1
Then y = –x
y = –(–1)
y = 1
Table
Points X Y Z
x 1 2 –1
y – 1 – 2 1
Coordinates (1, –1) (2,–2) ( –1,1)
Reflection
All lines pass through the origin.
Since a is negative, all lines fall from left to right. Greater magnitude of a gives a steeper line.
✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️
(iii) Graphs of
y = 5x ..........(1)
y = –5x ..........(2)
From equation (1)
y = 5x
(1) If x = 1
Then y = 5x
y = 5 × (1)
y = 5
(2) If x = 2
Then y = 5x
y = 5 × (2)
y = 10
(3) If x = –1
Then y = 5x
y = 5 × (–1)
y = –5
Table
Points A B C
x 1 2 –1
y 5 10 – 5
Coordinates (1,5) (2,10) ( –1,–5)
From equation (2)
y = –5x
(1) If x = 1
Then y = –5x
y = –5 × (1)
y = –5
(2) If x = 2
Then y = –5x
y = –5 × (2)
y = –10
(3) If x = –1
Then y = –5x
y = –5 × (–1)
y = 5
Table
Points A B C
x 1 2 –1
y –5 –10 5
Coordinates (1,–5) (2,–10) ( –1,5)
✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️
(iv) Graphs of
y = 3x – 1 ..........(1)
y = 3x + 1 ..........(2)
y = 3x ..........(3)
From equation (1)
y = 3x – 1
(1) If x = 1
Then y = 3x – 1
y = 3(1) – (1)
y = 3 –1
y = 2
(2) If x = 2
Then y = 3x – 1
y = 3(2) – (1)
y = 6 –1
y = 5
(3) If x = 3
Then y = 3x – 1
y = 3(3) – (1)
y = 9 –1
y = 8
(4) If x = –1
Then y = 3x – 1
y = 3(–1) – (1)
y = –3 –1
y = – 4
Table
Points A B C D
x 1 2 3 –1
y 2 5 8 –4
Coordinates (1,2) (2,5) ( 3,8) (–1,–4)
From equation (2)
y = 3x + 1
(1) If x = 1
Then y = 3x + 1
y = 3(1) + (1)
y = 3 +1
y = 4
(2) If x = 2
Then y = 3x + 1
y = 3(2) + (1)
y = 6 +1
y = 7
(3) If x = 3
Then y = 3x + 1
y = 3(3) + (1)
y = 9 +1
y = 10
(4) If x = –1
Then y = 3x + 1
y = 3(–1) + (1)
y = –3 +1
y = – 2
Table
Points A B C D
x 1 2 3 –1
y 4 7 10 –2
Coordinates (1,4) (2,7) ( 3,10) (–1,–2)
From equation (3)
y = 3x
(1) If x = 1
Then y = 3x
y = 3(1)
y = 3
(2) If x = 2
Then y = 3x
y = 3(2)
y = 6
(3) If x = 3
Then y = 3x
y = 3(3)
y = 9
(4) If x = –1
Then y = 3x
y = 3(–1)
y = –3
Table
Points A B C D
x 1 2 3 –1
y 3 6 9 –3
Coordinates (1,4) (2,7) ( 3,10) (–1,–2)
✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️
(v) Graphs of
y = 2x – 3 ..........(1)
y = 2x + 3 ..........(2)
y = –2x ..........(3)
From equation (1)
y = 2x – 3
(1) If x = 1
Then y = 2x – 3
y = 2(1) – 3
y = 2 –3
y = –1
(2) If x = 2
Then y = 2x – 3
y = 2(2) – 3
y = 4 –3
y = 1
(3) If x = –1
Then y = 2x – 3
y = 2(–1) – 3
y = –2 –3
y = –5
Table
Points A B C
x 1 2 –1
y –1 1 –5
Coordinates (1,2) (2,5) ( 3,8)
From equation (2)
y = 2x + 3
(1) If x = 1
Then y = 2x + 3
y = 2(1) + 3
y = 2 + 3
y = 5
(2) If x = 2
Then y = 2x + 3
y = 2(2) + 3
y = 4 + 3
y = 7
(3) If x = –1
Then y = 2x + 3
y = 2(–1) + 3
y = –2 + 3
y = 1
Table
Points A B C
x 1 2 –1
y 5 7 1
Coordinates (1,5) (2,7) (–1,1)
From equation (3)
y = – 2x
(1) If x = 1
Then y = – 2x
y = – 2(1)
y = – 2
(2) If x = 2
Then y = –2x
y = –2(2)
y = –4
(3) If x = –1
Then y = –2x
y = –2(–1)
y = 2
Table
Points A B C
x 1 2 –1
y – 2 –4 2
Coordinates (1,–2) (2,–4) (–1,1)
END-OF-CHAPTER EXERCISES
1. Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is –7.
Sol.
A polynomial of degree 3 is
x³ – 7x² – 5x + 1
Hare, the coefficient of x² is 7
Q. 2. Find the values of the following polynomials at the indicated values of the variables.
(i) 5x² – 3x + 7 if x = 1
(ii) 4t³ – t² + 6 if t = a
Sol.
(i) 5x² – 3x + 7
if x = 1
5x² – 3x + 7
= 5(1)² – 3(1) + 7
= 5×1 – 3 + 7
= 5 + 4
= 9
(ii) 4t³ – t² + 6
if t = a
4t³ – t² + 6
= 4(a)³ – (a)² + 6
= 4a³ – a² + 6
Q. 3. If we multiply a number by 5/2 and 2/3 add to the product, we get –7/12. Find the number.
Sol.
Let the number is = x
A.T.Q.
(5/2)x + 2/3 = –7/12
To remove denominator Multiply by the LCM of 2, 3 and 12. Which is 12
[(5/2)x + 2/3 = –7/12] × 12
12 × 5/2 × x + 12× 2/3 = 12× –7/12
30x + 8 = –7
30x = –7–8
x = –15/30
x = 1/2
Q. 4. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Sol.
Let the smaller number be x.
Then the larger number is = 5x
After acding 21.
New numbers ar
the smaller number be x +21.
Then the larger number is = 5x + 21
According to the question A.T.Q.
Big No. = Small No.
5x + 21 = 2(x +21)
5x + 21 = 2x + 42
5x – 2x = 42 – 21
3x = 21
x = 21/3
x = 7
Small = 7
Big No. = 5 × 7
= 35
Q. 5. If you have ₹800 and you save ₹250 every month, find the amount you have after
(i) 6 months
(ii) 2 years.
Express this as a linear pattern.
Sol.
Initial amount = ₹ 800
Monthly saving = ₹ 250
Linear expression after 'n' manthes
= 800 + 250 n
(i) After 6 months
= 800 + 250 n
= 800 + 250(6)
= 800 + 1500
= ₹2300
(ii) After 2 years
2 years = 24 months
= 800 + 250 n
= 800 + 250(24)
= 800 + 6000
= ₹6800
*Q. 6. The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers. he original
Sol.
Let the tens digit be x
Then the ones digit is x – 3
Onginal number:
= 10x + (x –3)
= 11x – 3
Now Interchanges the number:
= 10(x –3) + x
= 10x – 30 + x
= 11x –30
According to the question (A.T.Q.)
(11x – 3) + (11x – 30) = 143
11x – 3 + 11x – 30 = 143
22x – 33 = 143
22x = 143 + 33
x = 176/22
x = 8
Tens digit = 8
Ones digit = 8 –3
Ones digit 5
So the numbers are = 85 and 58
🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺
For a line in the form:
y = mx + c
m = slope
c = y-intercept
The line cuts the y-axis at (0, c)
🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺
*Q. 7. Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis.
(i) y = – 3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x – 10
(iv) 3y = 6x – 11
Are any of the lines parallel?
Sol.
For equation (1)
y = – 3x + 4
(1) If x = 1
Then y = – 3x + 4
y = – 3(1) + 4
y = –3 + 4
y = 1
(2) If x = 2
Then y = – 3x + 4
y = – 3(2) + 4
y = –6 + 4
y = –2
(3) If x = – 1
Then y = – 3x + 4
y = – 3(–1) + 4
y = 3 + 4
y = 7
Table
Points A B C
x 1 2 –1
y 1 –2 7
Coordinates (1,1) (2,–2) (–1,7)
(i) y = – 3x + 4
Comparing with y = mx + c
Slope m = –3
y-intercept c = 4
So, the line cuts the y-axis at (0,c) : (0,4)
✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️
For equation (2)
2y = 4x + 7
(1) If x = 1
Then 2y = 4x + 7
2y = 4(1) + 7
2y = 4 + 7
y = 11/2
(2) If x = 2
Then 2y = 4x + 7
2y = 4(2) + 7
2y = 8 + 7
y = 15/2
(3) If x = –1
Then 2y = 4x + 7
2y = 4(–1) + 7
2y = –4 + 7
y = 3/2
Table
Points A B C
x 1 2 –1
y 11/2 15/2 3/2
Coordinates (1,11/2) (2,15/2) (–1,3/2)
(ii) 2y = 4x + 7
[ 2y = 4x + 7] ÷ 2
y = 2x + 7/2
Comparing with y = mx + c
Slope m = 2
y-intercept c = 7/2
So, the line cuts the y-axis at (0,c) : (0,7/2)
✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️
For equation (3)
5y = 6x – 10
(1) If x = 1
Then 5y = 6x – 10
5y = 6(1) – 10
5y = 6 – 10
y = –4/5
(1) If x = 2
Then 5y = 6x – 10
5y = 6(2) – 10
5y = 12 – 10
y = 2/5
(3) If x = 3
Then 5y = 6x – 10
5y = 6(3) – 10
5y = 18 – 10
y = 8/5
Table
Points A B C
x 1 2 3
y –4/5 2/5 8/5
Coordinates (1,–4/5) (2,2/5) (3,8/5)
(iii) 5y = 6x – 10
[ 5y = 6x – 10] ÷ 5
y = 6/5 x + 10/5
y = 6/5 x + 2
Now
Comparing with y = mx + c
Slope m = 6/5
y-intercept c = 2
So, the line cuts the y-axis at (0,c) : (0,2)
✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️
For equation (4)
3y = 6x – 11
(1) If x = 1
Then 3y = 6x – 11
3y = 6(1) – 11
3y = 6 – 11
y = –5/3
(2) If x = 2
Then 3y = 6x – 11
3y = 6(2) – 11
3y = 12 – 11
y = 1/3
(3) If x = 3
Then 3y = 6x – 11
3y = 6(3) – 11
3y = 18 – 11
y = 7/3
Table
Points A B C
x 1 2 3
y –5/3 1/3 7/3
Coordinates (1,–5/3) (2,1/3) (3,7/3)
(iii) 3y = 6x – 11
[ 3y = 6x – 11] ÷ 3
y = 6/3 x – 11/3
y = 2 x – 11/3
Now
Comparing with y = mx + c
Slope m = 2
y-intercept c = 11/3
So, the line cuts the y-axis at (0,c) : (0,11/3)
✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️
*Q. 8. If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y °F, the relation between the two systems of measurement of temperature is given by the linear equation y = 9/5 (x – 273) + 32
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 Κ.
(ii) If the temperature is 158 °F, then find the temperature in Kelvin.
Sol.
The relation between Kelvin and Fahrenheit is:
y = 9/5 (x – 273) + 32
where
x = Ternaerature in Kelvin
y = temperature in Fahrenheit
(1) Find Fahrenheit temperature
when x = 313K
Substitute x = 313
y = 9/5 (313 – 273) + 32
= 9/5 (40) + 32
= 72 + 32
= 104
(ii) Find Kelvin temperature when y = 158 °F 158 = 9/5(x – 273) + 32
9/5(x – 273) + 32 = 158
9/5(x – 273) = 158 – 32
9/5(x – 273) = 126
(x – 273) = 126 × 5/9
(x – 273) = 70
x = 70 + 273
x = 303 K
*Q. 9. The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force. Express this in the form of a linear equation in two variables (work w and distance d), and draw its graph by taking the constant force as 3 units. What is the work done when the distance travelled is 2 units? Verify it by plotting it on the graph.
Sol.
We know that
Work Done = Force × Distance
Let:
Work done = w
Distance = d
Given constant force = 3 units
Sol
w = 3d
(This is the required linear equation)
Draw the graph
Take some values of d
Distance (d). Work 3d
0 0
1 3
2 6
3 9
4 12
Plot the points:
(0,0), (1,3), (2,6), (3,9) and join them to get a straight line.
Work done when distance = 2 units
= 3(2) unit
= 6
*Q. 10. The graph of a linear polynomial p(x) passes through the points
(1, 5) and (3, 11).
(i) Find the polynomial p(x)
(ii) Find the coordinates where the graph of p(x) cuts the axes.
(iii) Draw the graph of p(x) and verify your answers.
Sol.
The graph aasses through: (1,5) ancl (3,11)
Lel.
P(x) = ax + c
1) Find equation p(x)
Using point (1, 5)
a(1) + b = 5
a + b = 5 ..........(1)
Using point (3, 11)
a(3) + b = 11
3a + b = 11 ..........(1)
Subtract (1) free (2):
(3a + b) – (a + b) = 11 – 5
3a + b – a – b = 11 – 5
2a = 6
a = 6/2
a = 3
Substitute a = 3 into equation (1):
a + b = 5
3 + b = 5
b = 5 –3
b = 2
Therefors,
p(x) = 3a + 2
Then
y = 3a + 2
(ii) Coordinates where graph cuts the y-axis
Put x = 0
y = 3(0) + 2
y = 2
Sol that Point: (0,2)
Coordinates where graph cuts the x-axis
Put y = 0
0 = 3x + 2
3x = – 2
x = –2/3
Point: (–2/3, 0)
iii)
For equation (4)
y = 3x + 2
(1) If x = 1
Then y = 3x + 2
y = 3(1) + 2
y = 3 + 2
y = 5
(2) If x = 2
Then y = 3x + 2
y = 3(2) + 2
y = 6 + 2
y = 8
(1) If x = 3
Then y = 3x + 2
y = 3(3) + 2
y = 9 + 2
y = 11
Table
Points A B C
x 1 2 3
y 5 8 11
Coordinates (1,5) (2, 8) (3, 11)
Draw the graph
Plot these points:
(1,5), (2,8), (3,11)
Join them with a straight line.
The line cuts:
y-axis at (0, 2)
⚫x-axis at (–2/3, 0)
Hence verified.
*Q. 11. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) p(0) = 5
(ii) The polynomial p(x) – q(x) cuts the x-axis at (3, 0).
(iii) The sum p(x) + q(x) is equal to 6x + 4 for all real shed
Find the polynomials p(x) and q(x)
Sol.
Given
p(x) = ax + b
q(x) = cx + d
For part (i)
p(0) = 5
a(0) + b = 5
0 + b = 5
b = 5
Now
p(x) = ax + b
p(x) = ax + 5
For part (ii)
p(x) + q(x) = 6x + 4
(ax + 5) + (cx + d) = 6x + 4
ax + 5 + cx + d)= 6x + 4
ax + cx + 5 + d = 6x + 4
(a + c)x + (5 + d) = 6x + 4
Comparing both sides
(a + c)x = 6x and (5 + d) = 4
a + c = 6 and d = 4 –5 = –1
a + c = 6 ...........(1)
and d = –1
q(x) = cx + d
q(x) = cx – 1
For part (iii)
p(x) – q(x) = (ax + 5) – (cx –1)
= ax + 5 – cx + 1
= ax – cx + 5 + 1
= (a – c)x + 6
The polynomial p(x) – q(x) cuts the x-axis at (3, 0).
(a – c)3 + 6 = 0
(a – c)3 = –6
(a – c) = – 6 /3
(a – c) = – 2 ...........(2)
By solving the equation (1) and (2)
a + c = 6
a – c = –2
2.a = 4
a = 4/2
a = 2
Substituting a = 2 in equation (1)
a + c = 6
2 + c = 6
c = 6 – 2
c = 4
Final polynomial
p(x) = 2x + 5
g(x) = 4x – 1
*Q. 12. Look at the first three stages of a growing pattern of hexagons made using matchsticks. A new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage.
(i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages?
(ii) Complete the following table.
(iii) Find a rule to determine the number of matchsticks required for the n(th) stage. Introduction to Linear Polynomials
(iv) How many matchsticks will be required for the 15th stage of the pattern?
(v) Can 200 matchsticks form a stage in this pattern? Justify your answer.
Sol.
From the figures:
Stage 1 has 1 hexagon
Stage 2 has 2 joined hexagons
Stage 3 has 3 joined hexagons
We find that Each new hexagon shares one side with the previous hexagon.
A hexagon normally needs 6 matchsticks, but when one side is shared, only 5 extra matchsticks are added.
6 + 5×0 = 6 6 + 5×1 =11 6 + 5×2 =16
Now formula = 6 + 5 × (n –1)
(i) Next two stages and number of matchsticks
Stage 4
n = 4
Number of stocks = 6 + 5 × (4–1)
= 6 + 5×3
= 6 + 15
= 21
So, Stage 4 needs: 21 matchsticks
Stanumber of matchsticks for Stage 5
n = 5
Number of stocks = 6 + 5 × (5–1)
= 6 + 5×4
= 6 + 20
= 26
So, Stage 5 needs: 26 matchsticks
(ii) Complete the table
Stage Number. 1 2 3 4 5
No. of sticks 6 11 16 21 26
For 6th = Number of stocks = 6 + 5 × (6–1)
= 6 + 5×5
= 6 + 25
= 31
So, Stage 6 needs: 31 matchsticks
For nth = Number of stocks = 6 + 5 × (n–1)
(iii) Rule for the nth stage
Number of stocks for nth stage
= 6 + 5 × (n–1)
= 6 + 5 n – 5
for nth stage = 5n +1
(iv) Matchsticks required for the 15th stage
Using the rule: M = 5n +1
Put n = 15
M = 5 (15) + 1
= 75 + 1
= 76
76 Matchsticks required for the 15th stage
(v) Can 200 matchsticks form a stage?
Use the rule: M = 5n +1
5n +1 = 200
5n = 199
n = 199
n = 39.8
But stage number must be a whole number.
So, 200 matchsticks cannot form any stage in this pattern.
*Q. 13. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) The graph of p(x) passes through the points (2, 3) and (6, 11).
(ii) The graph of q(x) passes through the point (4, -1).
(iii) The graph of q(x) is parallel to the graph of p(x)
Find the polynomials p(x) and q(x) Also, find the coordinates of the point where these lines meet the x-axis.
Sol.
13. Find the polynomials p(x) and q(x)
Given:
p(x) = ax + b
q(x) = cx + d
Step 1:
Find p(x)
The graph of p(x) passes through:
(2,3) and (6, 11)
y2 – y1
First find the slope m = ________
x2 – x1
11 – 3
First find the slope m = ________
6 – 2
= 8/4
= 2 ........(1)
So,
p(x) = 2x + b
Use point (2,3):
3 = 2(2) + b
3 = 4 + b
b = –1
Therefore,
p(x) = 2x – 1
y = 2x – 1
Step 2:
Find q(x)
The graph of q(x) is parallel to p(x).
Parallel lines have the same slope.
So slope of q(x) is also 2. [From equation (1)]
Hence, q(x) = 2x + d
Given that q(x) passes through (4,–1):
–1 = 2 (4) + d
– 1 = 8 + d
d = –9
Therefore,
q(x) = 2x – 9
y = 2x – 9
Step 3:
Find where the lines meet the x-axis
A line meets the x-axis where: y = 0
For
p (x) = 2x – 1
0 = 2x –1
2x = 1
x = 1/2
Point: (1/2, 0)
For
q(x) = 2x – 9
0 = 2x – 9
2x = 9
x = 9/2
Point: (9/2,0)
Final Answers
p (x) = 2x – 1
q(x) = 2x – 9
*Q. 14. What do all linear functions of the form f(x) = ax + a, a > 0 have in common?
Sol.
Given:
f(x) = ax + a
Common property of all functions
f(x) = ax + a, a > 0
Take common factor a:
f(x) = a(x+1)
Since a > 0:
The slope is positive.
Therefore, all graphs rise from left to right.
Now find where the graph cuts the x-axis.
Put f(x) = 0:
a(x+1) = 0
Since a ≠ 0,
x + 1=0
x = –1
So every graph passes through the point:
(–1, 0)
Common property:
All graphs of the form
with a > 0
f(x) = ax + a
• are increasing straight lines, and
• pass through the fixed point
(–1, 0)
Question 14.
What do all linear functions of the form f(x) = ax + a, a > 0, have in common?
Solution:
Given: f(x) = ax + a, where a > 0
Common properties of all such linear functions are:
1. Slope:
The slope is a, and since a > 0 all the lines have positive slope. So, all these lines rise from left to right.
2. y-intercept:
Putting x = 0
f(0) = a So, the y-intercept is (0, a).
Since a > 0 all the lines cut they-axis above the origin.
3. x-intercept:
ax + a = 0
⇒ a(x + 1) = 0
To find where the line cuts the x-axis, put f(x) = 0, Since a > 0, –
a is not zero.
So, x + 1 = 0
⇒ x = –1
Thus, every line cuts the x-axis at the same point (–1,0).
Therefore, all linear functions of the form f(x) = ax + a, a > 0 have the following common characteristic:
1. All have positive slope.⁹ó
2. All cut the y-axis above the origin.
3. All pass through the fixed point (-1, 0).
CHAPTER SUMMARY
An algebraic expression combines numbers, variables, and operation symbols. For example, 2x² + 5xy - 3y² is an algebraic expression in the variables x and y. 2x², 5xy and3y² are the terms of the algebraic expression, and the numbers 2, 5 and -3 are coefficients of the terms.
Univariate Polynomials are algebraic expressions in one variable. Thus x² + 5x + 3 and 3y³ - 4y² + 5 are univariate polynomials in x and y, respectively. The highest power of the variable in a univariate polynomial is called its degree. Thus, x² + 5x + 3 is of degree 2 while 3y³ - 4y² + 5 is of degree 3.
A polynomial of degree one is called a linear polynomial. Hence, 2x + 3 and 5-4y are linear polynomials in the variables x and y, respectively.
Linear growth refers to a pattern in which a quantity increases by a fixed amount over equal intervals. In contrast, linear decay describes a pattern in which a quantity decreases by a fixed amount over equal intervals.
A linear pattern is a sequence of numbers where the difference between consecutive terms is constant.
A linear relationship between two variables x and y is represented by a straight line y = ax + b. The slope of this line is a. The constant b is called the y-intercept which is the distance from the origin where the line cuts the y-axis. When b = 0 the equation of the line becomes y = ax and the line passes through the origin.
Linear growth is represented by a straight line with positive slope and linear decay is represented by a straight line with negative slope.
Parallel lines are of the form y = ax + b. where the slope a is fi while b, the y-intercept, varies.
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