Class 10 Exercise-3.3
Q1. Solve the following pair of linear equation by the substitution method
(i) x + y = 14
x - y = 4
Solution:
x + y = 14 ..........(1)
x - y = 4 .........(2)
You can use one of the equations to solve this question
x - y = 4
x = 4 + y ..
Putting the value of x in equation (1)
(4+y) + y = 14
4 + 2y = 14
2y = 14-2
2y = 10
y = ${10}/2$
y = 5
Putting value of y in equation (2)
x - 5 = 4
x = 4+5
x = 9
Therefore, x = 9 and y = 5
(ii) s - t = 3
$s/3$ + $t/2$ = 6
Solution:
Given, s - t = 3 ----------------- (1)
$s/3$ + $t/2$ = 6 -------------
(2)
s - t = 3
s = 3 + t
Putting the value of s in equation 2
${3 + t}/3 + t/2 = 6$
${2(3+t) + 3t}/6 = 6$
$6 + 2t + 3t = 36$
$6 + 5t = 36$
5t = 36-6
5t = 30
t = ${30}/5$
t = 6
Putting value of t in equation (1)
s - 6 = 3
s = 3+ 6= 9
Therefore, s = 9 and t = 6