Question

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

Answer 1

Answer:

7, 9

Step-by-step 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.