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Re: There are two positive integers a and b, where a > b. [#permalink]
Archit3110 wrote:
GMATNinja wrote:
There are two positive integers \(𝑎\) and \(𝑏\), where \(𝑎 > 𝑏\). The sum of the squares of these integers is equal to 5 times their sum. The difference of the squares of these integers is equal to 9 times their difference. What is the value of \(𝑎 − 𝑏\)?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7


given
a>b
a^2+b^2 = 5*( a+b)
a^2-b^2= 9* (a-b)
add both
2*a^2 = 14a-4b
a^2 = 7a-2b
plugin values we see at a = 6 and b = 3 expression is valid
a-b= 3 correct option a


hi
In this question how did you plug in the value? I didn't understand the last step can you please explain that again
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Re: There are two positive integers a and b, where a > b. [#permalink]
Please explain how you plugged in the values

Posted from my mobile device
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Re: There are two positive integers a and b, where a > b. [#permalink]
1
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we have:

1. a^2 + b^2 = 5 ( a + b )
2. a^2 - b^2 = 9 ( a - b )

If we expand 2. we will have ( a + b )( a - b ) = 9 ( a - b )
we can cancel out (a-b) from both sides, then we will have (a+b) = 9

Replace it in 1. we will have a^2 + b^2 = 5 (9) = 45

Also a>b, so the possible combinations are either a=5,b=4 / a=6,b=3 / a=7,b=2 / a=7,b=1

See each one that satisfies both equations a+b = 9 and a^2 + b^2 = 45

We can see that 6^2+3^2=45 and 6+3=9.
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Re: There are two positive integers a and b, where a > b. [#permalink]
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