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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:
[tex]x=2y-5[/tex]
We also know that the sum of the squares of the two positive integers is 130. Using this information, we can write another equation.
[tex]x^2+y^2=130[/tex]
These two equations form a system of equations.
[tex]x=2y-5[/tex]
[tex]x^2+y^2=130[/tex]
We can solve this system of equations by the substitution method. Using this method, we substitute [tex]2y-5[/tex] for [tex]x[/tex].
[tex](2y-5)^2+y^2=130[/tex]
Expand [tex](2y-5)^2[/tex] . Notice that [tex](a-b)^2=a^2-2ab+b^2[/tex] :
[tex](2y-5)^2=4y^2-20y+25[/tex]
[tex]4y^2+y^2-20y+25=130[/tex]
Subtract 25 from both sides
[tex]4y^2+y^2-20y=105[/tex]
Add like terms
[tex]5y^2-20y=105[/tex]
Divide both sides by 5
[tex]y^2-4y=21[/tex]
Subtract 21 from both sides
[tex]y^2-4y-21 = 0[/tex]
Factor the equation
[tex]y^2-7y+3y-21=0\\y(y-7)+3(y-7)=0\\(y-7)(y+3)=0\\y=7\text{ or} \ -3[/tex]
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.
[tex]x=2y-5\\x=2(7)-5\\x=14-5\\x=9[/tex]
[tex]CHECK:\\x^2+y^2=7^2+9^2=49+81=130 \Rightarrow Correct![/tex]
Therefore the two integers are 7 and 9.