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 $9$.