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

Given :

The sum of two numbers is 8.

Their sum is four times their difference.

To find :

We have to find the numbers.

Solution :

Let the numbers be x and y.

This implies,

$x+y=8$

$x = 8-y$.....(i)

The sum is four times their difference.

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

$x+y=4x-4y$ or $x+y=4y-4x$

$y+4y=4x-x$ or $x+4x=4y-y$

$5y=3x$ or $5x=3y$

Substituting $x = 8-y$, we get,

$5y = 3(8-y)$ or $5(8-y)=3y$

$5y = 24-3y$ or $40-5y=3y$

$5y+3y=24$ or $3y+5y=40$

$8y=24$ or $8y=40$

$y =\frac{24}{3}$ or $y =\frac{40}{8}$

$y = 3$ or $y=5$

If $y = 3$ then $x=8-3=5$

If $y=5$ then $x=8-5=3$

The required numbers are $5$ and $3$.

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Updated on: 10-Oct-2022

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