The sum of two numbers is 8. If their sum is four times their difference, find the numbers.

Solution:

Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.

The sum of the two numbers is 8. Thus, we have 

The sum of the two numbers is four times their difference. Thus, we have

$x+y=4(x-y)$

$\Rightarrow x+y=4 x-4 y$

$\Rightarrow 4 x-4 y-x-y=0$

$\Rightarrow 3 x-5 y=0$

So, we have two equations

$x+y=8$

 

$3 x-5 y=0$

Here x and y are unknowns. We have to solve the above equations for x and y.

Multiplying the first equation by 5 and then adding with the second equation, we have

$5(x+y)+(3 x-5 y)=5 \times 8+0$

$\Rightarrow 5 x+5 y+3 x-5 y=40$

$\Rightarrow 8 x=40$

$\Rightarrow x=\frac{40}{8}$

$\Rightarrow x=5$

Substituting the value of x in the first equation, we have

$5+y=8$

$\Rightarrow y=8-5$

$\Rightarrow y=3$

Hence, the numbers are 5 and 3.