## NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

### NCERT Solutions

#### Exercise 3.1

Question.1. Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
Solution: Let number of boys = x and number of girls = y
We have, x + y = 10 …(i)
and y = x + 4 …(ii)
Let us find some points of equation (i) to plot on the graph.
x + y = 10
y = 10 – x ……(iii)
Putting x = 4 in equation (iii), we get
y = 10 – 4 = 6
Putting x = 3 in equation (iii), we get
y = 10 – 3 = 7
Putting x = 5 in equation (iii), we get
y = 10 – 5 = 5
Therefore, Some points on line (i) are:

 x 4 3 5 y 6 7 5

Let us find some points of equation (ii) to plot on the graph.
y = x + 4
Putting x = 0 in equation (ii), we get
y = 0 + 4 = 4
Putting x = 3 in equation (ii), we get
y = 3 + 4 = 7
Putting x = 2 in equation (ii), we get
y = 2 + 4 = 6
Therefore, Some points on line (ii) are:
The two lines meet at A (3, 7), i.e., x = 3, y = 7
Hence, number of boys = 3, number of girls = 7.

 x 0 3 2 y 4 7 6

(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.
Solution: Let cost of one pencil = ₹ x
and cost of one pen = ₹ y
According to the given condition, we have
5 x + 7 y = 50 …(i)
and 7 x + 5 y = 46 …(ii)
Let us find some points of equation (i) to plot on the graph.
5 x + 7 y = 50
y = \frac{50-5x}{7} ……(iii)
Putting x = 10 in equation (iii), we get
y = \frac{50-5 \times 10}{7} = \frac{0}{7} = 0
Putting x = 3 in equation (iii), we get
y = \frac{50-5 \times 3}{7} = \frac{35}{7} = 5
Putting x = 0 in equation (iii), we get
y = \frac{50-5 \times 0}{7} = \frac{50}{7} = 7.14
Therefore, Some points on line (i) are:

 x 10 3 0 y 0 5 7.14

Let us find some points of equation (ii) to plot on the graph.
7 x + 5 y = 46
y = \frac{46-7x}{5} ……(iv)
Putting x = 0 in equation (iv), we get
y = \frac{46-7 \times 0}{5} = \frac{46}{5} = 9.2
Putting x = 5 in equation (iv), we get
y = \frac{46-7 \times 5}{5} = \frac{11}{5} = 2.2
Putting x = 3 in equation (iv), we get
y = \frac{50-7 \times 3}{7} = \frac{29}{7} = 4.14
Therefore, Some points on line (ii) are:

 x 0 5 3 y 9.2 2.2 4.14

Question.2. On comparing the ratios \frac{a_{1}}{a_{2}} , \frac{b_{1}}{b_{2}} and \frac{c_{1}}{c_{2}} , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5 x – 4 y + 8 = 0
7 x + 6 y – 9 = 0
Solution: The given pair of linear equation are :
5 x – 4 y + 8 = 0 …(i)
7 x + 6 y – 9 = 0 …(ii)
Comparing the eqn. (i) and (ii) with a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 respectively,
a_{1} = 5, b_{1} = – 4, c_{1} = 8
a_{2} = 7, b_{2} = 6, c_{2} = – 9
Here,
\frac{a_{1}}{a_{2}} = \frac{5}{7}
\frac{b_{1}}{b_{2}} = \frac{-4}{6} = \frac{-2}{3}
\frac{c_{1}}{c_{2}} = \frac{8}{-9}
Clearly, \frac{5}{7} \frac{-2}{3} \frac{8}{-9}
Hence, \frac{a_{1}}{a_{2}} \frac{b_{1}}{b_{2}} \frac{c_{1}}{c_{2}}
Thus, lines (i) and (ii) intersect at a point, that is, they are intersecting lines.

(ii) 9 x + 3 y + 12 = 0
18 x + 6 y + 24 = 0
Solution: The given pair of linear equation are :
9 x + 3 y + 12 = 0 …(i)
18 x + 6 y + 24 = 0 …(ii)
Comparing the eqn. (i) and (ii) with a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 respectively,
a_{1} = 9, b_{1} = 3, c_{1} = 12
a_{2} = 18, b_{2} = 6, c_{2} = 24
Here,
\frac{a_{1}}{a_{2}} = \frac{9}{18} = \frac{1}{2}
\frac{b_{1}}{b_{2}} = \frac{3}{6} = \frac{1}{2}
\frac{c_{1}}{c_{2}} = \frac{12}{24} = \frac{1}{2}
Clearly, \frac{1}{1} = \frac{1}{2} = \frac{1}{2}
Hence, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}
Hence, lines (i) and (ii) are coincident lines.

(iii) 6 x – 3 y + 10 = 0
2 x y + 9 = 0
Solution: The given pair of linear equation are :
6 x – 3 y + 10 = 0 …(i)
2 x y + 9 = 0 …(ii)
Comparing the eqn. (i) and (ii) with a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 respectively,
a_{1} = 6, b_{1} = – 3, c_{1} = 10
a_{2} = 2, b_{2} = – 1, c_{2} = 9
Here,
\frac{a_{1}}{a_{2}} = \frac{6}{2} = \frac{3}{1}
\frac{b_{1}}{b_{2}} = \frac{-3}{-1} = \frac{3}{1}
\frac{c_{1}}{c_{2}} = \frac{10}{9} = = \frac{10}{9}
Clearly, \frac{3}{1} = \frac{3}{1} \frac{10}{9}
Hence, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \frac{c_{1}}{c_{2}}
Thus, lines (i) and (ii) are parallel lines.

Question.3. On comparing the ratios \frac{a_{1}}{a_{2}} , \frac{b_{1}}{b_{2}} and \frac{c_{1}}{c_{2}} , find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3 x + 2 y = 5; 2 x – 3 y = 7
Solution: The given pair of linear equation are:
3 x + 2 y – 5 = 0 …(i)
2 x – 3 y – 7 = 0 …(ii)
Comparing the eqn. (i) and (ii) with a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 respectively,
a_{1} = 3, b_{1} = 2, c_{1} = – 5
a_{2} = 2, b_{2} = – 3, c_{2} = – 7
Here,
\frac{a_{1}}{a_{2}} = \frac{3}{2}
\frac{b_{1}}{b_{2}} = \frac{2}{-3}
\frac{c_{1}}{c_{2}} = \frac{-5}{-7} = \frac{5}{7}
Clearly, \frac{3}{2} \frac{2}{-3} \frac{5}{7}
Hence, \frac{a_{1}}{a_{2}} \frac{b_{1}}{b_{2}} \frac{c_{1}}{c_{2}}
So, pair of linear equations are consistent.

(ii) 2 x – 3 y = 8; 4 x – 6 y = 9
Solution: The given pair of linear equation are :
2 x – 3 y – 8 = 0 …(i)
4 x – 6 y – 9 = 0 …(ii)
Comparing the eqn. (i) and (ii) with a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 respectively,
a_{1} = 2, b_{1} = – 3, c_{1} = – 8
a_{2} = 4, b_{2} = – 6, c_{2} = – 9
Here,
\frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2}
\frac{b_{1}}{b_{2}} = \frac{-3}{-6} = \frac{1}{2}
\frac{c_{1}}{c_{2}} = \frac{-8}{-9} = \frac{8}{9}
Clearly, \frac{1}{2} = \frac{1}{2} \frac{8}{9}
Hence, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \frac{c_{1}}{c_{2}}
So, pair of linear equations are inconsistent.

(iii) \frac{3}{2}x + \frac{5}{3}y = 7; 9 x – 10 y = 14
Solution: The given pair of linear equation are :
\frac{3}{2}x + \frac{5}{3}y = 7
9 x + 10 y – 42 = 0 …(i)
9 x – 10 y – 14 = 0 …(ii)
Comparing the eqn. (i) and (ii) with a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 respectively,
a_{1} = 9, b_{1} = 10, c_{1} = – 42
a_{2} = 9, b_{2} = – 10, c_{2} = – 14
Here,
\frac{a_{1}}{a_{2}} = \frac{9}{9} = \frac{1}{1}
\frac{b_{1}}{b_{2}} = \frac{10}{-10} = \frac{1}{-1}
\frac{c_{1}}{c_{2}} = \frac{-42}{-14} = \frac{3}{1}
Clearly, \frac{1}{1} \frac{1}{-1} \frac{3}{1}
Hence, \frac{a_{1}}{a_{2}} \frac{b_{1}}{b_{2}} \frac{c_{1}}{c_{2}}
So, pair of linear equations are consistent.

(iv) 5 x – 3 y = 11; –10 x + 6 y = –22
Solution: The given pair of linear equation are :
5 x – 3 y – 11 = 0 …(i)
–10 x + 6 y + 22 = 0 …(ii)
Comparing the eqn. (i) and (ii) with a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 respectively,
a_{1} = 5, b_{1} = – 3, c_{1} = – 11
a_{2} = – 10, b_{2} = 6, c_{2} = 22
Here,
\frac{a_{1}}{a_{2}} = \frac{5}{-10} = \frac{1}{-2}
\frac{b_{1}}{b_{2}} = \frac{-3}{6} = \frac{-1}{2}
\frac{c_{1}}{c_{2}} = \frac{-11}{22} = \frac{-1}{2}
Clearly, \frac{-1}{2} = \frac{-1}{2} = \frac{-1}{2}
Hence, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}
So, pair of linear equations are consistent.

(v) \frac{4}{3}x + 2 y = 8; 2 x + 3 y = 12
Solution: The given pair of linear equation are :
\frac{4}{3}x + 2 y = 8
4 x + 6 y – 24 = 0 …(i)
2 x + 3 y – 12 = 0 …(ii)
Comparing the eqn. (i) and (ii) with a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 respectively,
a_{1} = 4, b_{1} = 6, c_{1} = – 24
a_{2} = 2, b_{2} = 3, c_{2} = – 12
Here,
\frac{a_{1}}{a_{2}} = \frac{4}{2} = \frac{2}{1}
\frac{b_{1}}{b_{2}} = \frac{6}{3} = \frac{2}{1}
\frac{c_{1}}{c_{2}} = \frac{-24}{-12} = \frac{2}{1}
Clearly, \frac{1}{1} = \frac{1}{-1} = \frac{3}{1}
Hence, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}
So, pair of linear equations are consistent.

##### Exercise 2.2

Question.1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients :
(i) x^{2}-2x-8
Solution: Let p(x) = x^{2}-2x-8
= x^{2}-4x+2x-8
= x(x-4)+2(x-4)
= (x – 4)(x + 2)
For finding the zeroes
Put, p(x) = 0
= (x – 4)(x + 2) = 0
x = 4, – 2

Verification: Now, sum of zeroes = 4 + (– 2)
= {\frac{2}{1}}
= {\frac{-(Coefficient  of  x)}{(Coefficient  of  x^{2})}}
Product of zeroes = 4 \times (– 2)
= {\frac{-8}{1}}
= {\frac{Constant  term}{Coefficient  of  x^{2}}}
Hence, relationship between the zeroes and the coefficients is true.

(ii) 4s^{2}-4s+1
Solution: Let p(s) = 4s^{2}-4s+1
= 4s^{2}-2s-2s+1
= 2s(2s-1)-1(2s-1)
= (2s – 1) (2s – 1)
= (2s-1)^{2}
For finding the zeroes, put p(s) = 0.
2s – 1 = 0
s = \frac{1}{2}
Hence, zeroes of p(s) are \frac{1}{2} and 12 \frac{1}{2} .

Verification: Now, sum of zeroes = \frac{1}{2} + \frac{1}{2}
= 1 = \frac{4}{4} = {\frac{-(Coefficient  of  s)}{(Coefficient  of  s^{2})}}
Product of zeroes = \frac{1}{2} \times \frac{1}{2}
= \frac{1}{4}
= {\frac{Constant  term}{Coefficient  of  s^{2}}}
Hence, relationship between the zeroes and the coefficients is true.

(iii) 6x^{2}-3-7x
Solution: Let p(x) = 6x^{2}-3-7x = 6x^{2}-7x-3
= 6x^{2}-9x+2x-3
= 3x(2x-3)+1(2x-3)
= (2x – 3) (3x + 1)
For finding the zeroes put p(x) = 0
(2x – 3) (3x + 1) = 0
x = {\frac{3}{2}}, {\frac{-1}{3}}
Hence, zeroes of the polynomial are {\frac{3}{2}} and {\frac{-1}{3}}.

Verification: Now sum of zeroes = {\frac{3}{2}} + {\frac{-1}{3}}
= {\frac{9-2}{6}}
= {\frac{7}{6}}
= {\frac{-(Coefficient  of  x)}{(Coefficient  of  x^{2})}}
Product of zeroes = {\frac{3}{2}} \times {\frac{(-1)}{3}}
= {\frac{-3}{6}}
= {\frac{Constant  term}{Coefficient  of  x^{2}}}

(iv) 4u^{2}+8u
Solution: Let p(u) = 4u^{2}+8u
= 4u (u + 2)
For finding the zeroes put p(u) = 0,
4u (u + 2) = 0
u = 0, – 2
Hence, zeroes are 0 and – 2.

Verification: Now sum of zeroes = 0 + (– 2)
= {\frac{-2}{1}} \times {\frac{4}{4}}
= {\frac{-8}{4}}
= {\frac{-(Coefficient  of  u)}{(Coefficient  of  u^{2})}}
Product of zeroes = 0 \times (– 2)
= {\frac{0}{1}} \times {\frac{4}{4}}
= {\frac{0}{4}}
= {\frac{Constant  term}{Coefficient  of  u^{2}}}
Hence, relationship between the zeroes and the coefficients is true.

(v) t^{2}-15
Solution: Let p(t) = t^{2}-15 .
= t^{2}-(\sqrt{15})^{2}
= (t-\sqrt{15})(t+\sqrt{15}) [∵ a^{2}-b^{2} = (a + b) (a – b)]
For finding the zeroes put p(t) = 0,
(t-\sqrt{15})(t+\sqrt{15}) = 0
t = -\sqrt{15}) , \sqrt{15}

Verification: Sum of the zeroes = -\sqrt{15}) + \sqrt{15}
= {\frac{0}{1}}
= = {\frac{-(Coefficient  of  t)}{(Coefficient  of  t^{2})}}
Product of zeroes = -\sqrt{15}) \times \sqrt{15}
= {-\frac{15}{1}}
= {\frac{Constant  term}{Coefficient  of  t^{2}}}
Hence, relationship between the zeroes and the coefficients is true.

(vi) 3x^{2}-x-4
Solution: Let p(x) = 3x^{2}-x-4
= 3x^{2}-4x+3x-4
= x(3x – 4) + 1 (3x – 4)
= (x + 1) (3x – 4)
For finding the zeroes put p(x) = 0,
(x + 1) (3x – 4) = 0
x = -1, {\frac{4}{3}}

Verification: Sum of zeroes = -1 + {\frac{4}{3}}
= {\frac{1}{3}}
= = {\frac{-(Coefficient  of  x)}{(Coefficient  of  x^{2})}}
Product of zeroes = -1 \times {\frac{4}{3}}
= {\frac{-4}{3}}
= {\frac{Constant  term}{Coefficient  of  x^{2}}}
Hence, relationship between the zeroes and the coefficients is true.

Question.2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) {\frac{1}{4}}, – 1
Solution: Let quadratic polynomial is p(x) whose zeroes are α and β.
Given that : α + β = 14 and αβ = – 1
So, required polynomial = k[x^{2} – (α + β)x + αβ]
= k\left[x^{2}-\frac{1}{4}x-1\right]
If k = 4, then 4x^{2}-x-4 .

(ii) \sqrt{2}, {\frac{1}{3}}
Solution: Let quadratic polynomial is ax^{2}+bx+c whose zeroes are α and β.
Given that: α + β = \sqrt{2} and αβ = 13
So, required polynomial = k[x^{2} – (α + β)x + αβ]
= k\left[x^{2}-\sqrt{2}x+{\frac{1}{3}}\right]
If k = 3, then 3x^{2}-3 \sqrt{2}x+1 .

(iii) 0, \sqrt{5}
Solution: Let quadratic polynomial is ax^{2}+bx+c whose zeroes are α and β.
Given that: α + β = 0 and αβ = \sqrt{5}
Hence, required polynomial = k[x^{2} – (α + β)x + αβ]
= k\left[x^{2}-0.x+ \sqrt{5} \right]
= k\left[x^{2}+ \sqrt{5} \right]
If k = 1, then x^{2}+ \sqrt{5} .

(iv) 1, 1
Solution: Let quadratic polynomial is ax^{2}+bx+c whose zeroes are α and β.
Given that : α + β = 1 and αβ = 1
Hence, required polynomial = k[x^{2} – (α + β)x + αβ]
= k\left[x^{2}-x+1 \right]
If k = 1, then x^{2}-x+1 .

(v) {\frac{-1}{4}}, {\frac{1}{4}}
Solution: Let quadratic polynomial is ax^{2}+bx+c whose zeroes are α and β.
Given that: α + β = -{\frac{1}{3}} and αβ = {\frac{1}{3}}
So, required polynomial = k[x^{2} – (α + β)x + αβ]
= k\left[x^{2}+\frac{1}{4}x+\frac{1}{4} \right]
If k = 4, then 4x^{2}+x+1 .

(vi) 4, 1
Solution: Let quadratic polynomial is ax^{2}+bx+c whose zeroes are α and β.
Given that: α + β = 4 and αβ = 1
Hence, required polynomial = k[x^{2} – (α + β)x + αβ]
= k\left[x^{2}-4x+1 \right].