Question

Find two numbers such that the sum of thrice the first and the second is 142,and four times the first exceeds the second by 138.

Answer 1

Let the two numbers be x and y

According to the question,

$3x+y=142$.....(1)

And,

$4x-y=138$.....(2)

By adding eq(1) and eq(2)
We get

$3x+y+4x-y=142+138$

$7x=280$

$x=\frac{280}{7}=40$

Putting $x=40$ in eq(1)
We get,

$3\left(40\right)+y=142$

$120+y=142$

$y=142-120=22$

Hence,

The two numbers are 40 and 22