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

User Comintern
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6 votes

Answer:

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.

User Mbanda
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