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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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