Question

One positive integer is 5 less than twice another. The sum of their squares is 130. Find the integers.

Answer 1

7, 9

Explanation:

Let x and y represent the two positive integers.

We can write x in terms of y. The question states "one positive integer is 5 less than twice another". Let "one positive integer" represent x.

Therefore:

`x=2y-5`

We also know that the sum of the squares of the two positive integers is 130. Using this information, we can write another equation.

`x^2+y^2=130`

These two equations form a system of equations.

`x=2y-5`

`x^2+y^2=130`

We can solve this system of equations by the substitution method. Using this method, we substitute
`2y-5` for
`x`.

`(2y-5)^2+y^2=130`

Expand
`(2y-5)^2` . Notice that
`(a-b)^2=a^2-2ab+b^2` :

`(2y-5)^2=4y^2-20y+25`

`4y^2+y^2-20y+25=130`

Subtract 25 from both sides

`4y^2+y^2-20y=105`

Add like terms

`5y^2-20y=105`

Divide both sides by 5

`y^2-4y=21`

Subtract 21 from both sides

`y^2-4y-21 = 0`

Factor the equation

`y^2-7y+3y-21=0\\y(y-7)+3(y-7)=0\\(y-7)(y+3)=0\\y=7\text{ or} \ -3`

Since the question states that the two integers are positive, one of the integers is 7.

We can use this information to find the other integer.

`x=2y-5\\x=2(7)-5\\x=14-5\\x=9`

`CHECK:\\x^2+y^2=7^2+9^2=49+81=130 \Rightarrow Correct!`

Therefore the two integers are 7 and 9.