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The difference of squares of two numbers is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers.
Given:
The difference of squares of two numbers is 88.
The larger number is 5 less than twice the smaller number.
To do:
We have to find the numbers.
Solution:
Let the two numbers be $x$ and $y$ where $y$ is the larger number.
This implies,
.$y=2x-5$
According to the question,
$y^2-x^2=88$
$(2x-5)^2-x^2=88$
$4x^2-20x+25-x^2=88$
$3x^2-20x+25-88=0$
$3x^2-20x-63=0$
Solving for $x$ by factorization method, we get,
$3x^2-27x+7x-63=0$
$3x(x-9)+7(x-9)=0$
$(3x+7)(x-9)=0$
$3x+7=0$ or $x-9=0$
$3x=-7$ or $x=9$
Considering the positive value of $x$, we get,
$x=9$ then $y=2(9)-5=18-5=13$
The required numbers are $13$ and $9$.
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