Question

The sum of two numbers is 20. The difference between three times the first rumber and twice the second is 40. Find the two numbers.

Answer 1

Let the first number is x and second number is y.

According to given conditions:

The sum of two numbers is 20.

$x+y=20$

And The difference between three times the first rumber and twice the second is 40.

$3x-2y=40$

Now multiply equation 1 with 2 and add in second eqution;

$\begin{gathered} 2(x+y)+(3x-2y)=2(20)+40 \\ 2x+2y+3x-2y=40+40 \\ 5x=80 \\ x=16 \end{gathered}$

Now put the value of x in equation 1:

$\begin{gathered} 16+y=20 \\ y=4 \end{gathered}$

So the first number is x=16 and second number is y=4.

According to given co