NCERT Solutions for Pair of Linear Equations in Two Variables Chapter 3 Ex 3.2 Class 10 Maths has all the solutions to the questions provided in the NCERT Book of the latest edition.
Students are advised to practice all the questions to get good marks in the board examination.
| Textbook | NCERT |
| Class | 10 |
| Subject | Mathematics |
| Chapter | 3 |
| Exercise | 3.2 |
| Chapter Name | Pair of Linear Equations in Two Variables |
Class 10 Maths Chapter 3 Exercise 3.2 Pair of Linear Equations in Two Variables NCERT Solution
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x – y = 4
Answer
x+y=14
⇒ x+14-y
x – y =4 —- (ii)
Putting value of x in eq(ii) from eq(i)
14-y-y=4
⇒ 14-4=2y
⇒ 2y=10
⇒ y=5
Putting value of y in eq(i)
x=14-5
⇒ x=9
(ii) s – t = 3
\frac{s}{3}+\frac{t}{2}=6
Answer
s-t=3
⇒ s=3-t
\frac{s}{3}+\frac{t}{2}=6
Putting value of s in eq(ii) from eq(i)
\frac{3+t}{3}+\frac{t}{2}=6
⇒ \frac{6+2t+3t}{6}=6
⇒ 6+5t=6\times6
⇒ 5t=36-6
⇒ t=\frac{30}{5}
⇒ t=6
Putting value of t in eq(i)
s=3+6
⇒ s=9
(iii) 3x – y = 3
9x – 3y = 9
Answer
3x-y=3
y=3x-3
9x-3y=9
Putting value of y in eq(ii) from eq(i)
9x-3(3x-3)=9
⇒ 9x-9x+9=9
⇒ 9=9
\because
So, the equation have infinitely many solution
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
Answer
0.2x+0.3y=1.3
⇒ 0.2x=1.3-0.3y
⇒ x=\frac{1.3-0.3y}{0.2}
0.4x+0.5y=2.3
Putting value of x in eq(ii) from eq(i)
0.4\left(\frac{1.3-0.3y}{0.2}\right) + 0.5y = 2.3
⇒ 2(1.3-0.3y)+0.5y=2.3
⇒ 2.6-0.6y+0.5y=2.3
⇒ 2.6-2.3-0.1y=0
⇒ 0.3=0.1y
⇒ y=\frac{0.3}{0.1}
⇒ y=3
Putting value of y in eq(i)
x=\frac{1.3-0.3\times3}{0.2}
⇒ x=\frac{1.3-0.9}{0.2}
⇒ x=\frac{0.4}{0.2}
⇒ x=2
(v) \sqrt{2}x+\sqrt{3}y = 0
\sqrt{3}x-\sqrt{8}y=0
Answer
\sqrt{2}x+\sqrt{3}y=0
⇒ y=\frac{-\sqrt{2}x}{\sqrt{3}}
\sqrt{3}x-\sqrt{8}y=0
Putting value of y from eq(i) in eq(ii)
\sqrt{3}x-\sqrt{8}\left(\frac{-\sqrt{2}x}{\sqrt{3}}\right)=0
⇒ 3x+\sqrt{16}x=0
⇒ 3x+4x=0
⇒ 5x=0
⇒ x=0
Putting value of x in eq(i)
y=0
(vi) \frac{3x}{2}-\frac{5y}{3}=-2
\frac{x}{3}+\frac{y}{2}=\frac{13}{6}
Answer
\frac{x}{3}+\frac{y}{2}=\frac{13}{6}
⇒ \frac{2x+2y}{6}=\frac{13}{6}
⇒ 2x=13-3y
⇒ x=\frac{13-3y}{2}
\frac{3}{2}x – \frac{5}{3}y = -2
Putting value of x in eq(ii) from eq(i)
\frac{3}{2}\times\left(\frac{13-3y}{2}\right) -\frac{5}{3}y=-2
⇒ \frac{39-9y}{4}-\frac{5}{3}y=-2
⇒ \frac{3(39-9y)-20y}{12}=-2
⇒ 117-27y-20y=-24
⇒ 117+24=47y
⇒ 47y=141
⇒ y=3
Putting value of y in eq(i)
x=\frac{13-9}{2}
⇒ x=\frac{4}{2}
⇒ x=2
Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
Answer
2x+3y=11
⇒ 2x=11-3y
2x-4y=-24
Putting value of 2x from eq(i)
⇒ (11-3y)-4y=-24
⇒ 11+24=7y
⇒ 7y=35
⇒ y=5
Putting value of y in eq(i)
2x=11-15
⇒ 2x=-4
⇒ x=-2
\because
⇒ 5 = -2m + 3
⇒ 2m = -2
⇒ m = -1
Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
Answer
Let bigger no. be x
smaller no.= y
According to question(A/Q),
x-y=26
⇒ x=26+y
and
x=3y
Putting value of x from eq(i)
⇒ 26+y=3y
⇒ 26=2y
⇒ y=13
Putting value of y in eq(i),
x=26+13
⇒ x=39
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Answer
Let larger angle be x^{\circ}
smaller angle = y^{\circ}
\because x+y=180^{\circ}
⇒ y=180-x
A/Q,
x=y+18
⇒ x=180-x+18
⇒ 2x=198
⇒ x=99^{\circ}
Putting value of x in eq(i), we get
y=180-99
⇒ y=81^{\circ}
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.
Answer
Let cost of each bat be ₹x
cost of each ball = ₹y
A/Q,
7x+6y=3800
⇒ y=\frac{3800-7x}{6}
and
3x+5y=1750
Putting value of from eq(i)
⇒ 3x+5\left(\frac{3800-7x}{6}\right)=1750
⇒ \frac{18x+1900-35x}{6}=1750
⇒ -17x+19000=10500
⇒ 17x=8500
⇒ x= ₹500
Putting value of x in eq(i), we get
y=\frac{3800-3500}{6}
⇒ y=\frac{300}{6}
⇒ y= ₹50
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Answer
let the fired charge be ₹x
the charge per Km = ₹y
A/Q,
x+10y=105
⇒ x=105-10y
and
x+15y=155
Putting value of x in eq(ii)
105-10y+15y=155
⇒ 5y=50
⇒ y= ₹10
Putting value of y in eq(i)
x=105-100
⇒ x= ₹5
\therefore
=25\times10+5
=250+5
= ₹255
(v) A fraction becomes \frac{9}{11}
Answer
Let numerator be x
denominator = y
then, fraction = \frac{x}{y}
A/Q,
\frac{x+2}{y+2}=\frac{9}{11}
⇒ 11x+22=9y+18
⇒ 11x=9y+18-22
⇒ x=\frac{9y-4}{11}
and
\frac{x+3}{y+3}=\frac{5}{6}
⇒ 6x+18=5y+15
⇒ 6x+18-15=5y
⇒ 6x+3=5y
Putting value of x
⇒ 6\left(\frac{9y-4}{11}\right)+3=5y
⇒ \frac{54y-24+33}{11}=5y
⇒ 54y+9=55y
⇒ y=9
Putting value of y in eq(i)
$$x=\frac{9\times9-4}{11}
⇒ x=\frac{77}{11}
⇒ x=7
\therefore
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Answer
Let present age of Jacob be x years
Present age of Jacob’s son = y years
Five years hence,
Age of Jacob = (x+5)
Age of Jacob’s Son = (y+5) yrs
A/Q,
(x+5)=3(y+5)
⇒ x=3y+15-5
⇒ x=3y+10
Five years ago,
Age of Jacob = (x-5)
Age of Jacob’s son = (y-5)
A/Q,
(x-5)=7(y-5)
⇒ x-5=7y-35
Putting value of x from eq(i)
⇒ 3y+10-5=7y-35
⇒ 5+35=7y-3y
⇒ 40=4y
⇒ y=10 years
Putting y in eq(i)
x=3\times10+10
⇒ x=40
\therefore
Present age of Jacob’s son = y = 10 years
Hope NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2, helps you in solving problems. If you have any doubts, drop a comment below and we will get back to you.