## 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].