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If \((ax + by)^2 = (bx + ay)^2\), what is the value of \(a – b\) ?
(1) \(x^2 > y^2\)
(2) a and b are positive integers.
(1) \(x^2 > y^2\)
(2) a and b are positive integers.
20 Nov 2022, 13:34
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\((ax+by)^2 = (bx+ay)^2\)
\(a^2x^2 + b^2y^2 + 2axby = b^2x^2+a^2y^2 + 2axby\)
We can subtract 2axby from both the sides of the equation
\(a^2(x^2-y^2) = b^2(x^2-y^2)\)
\((x^2-y^2)(a^2-b^2) = 0\)
Statement 1
\(x^2 > y^2\)
So we can conclude that \((x^2-y^2)\neq{0}\)
Hence \((a^2-b^2) = 0\)
(a+b)(a-b) = 0
We don't know the value of a + b, hence we can't conclude if (a+b) = 0 or (a-b) = 0 or both. Eliminate A and D.
Statement 2
a and b are positive integers.
Insufficient information. We can't find value of a - b. Eliminate B.
Combined
From 1, we know (a+b)(a-b) = 0
From 2, we know that \((a + b)\neq{0}\). This is because a and b are positive, hence the sum must also be +ve.
Therefore (a-b) = 0
Option C
\(a^2x^2 + b^2y^2 + 2axby = b^2x^2+a^2y^2 + 2axby\)
We can subtract 2axby from both the sides of the equation
\(a^2(x^2-y^2) = b^2(x^2-y^2)\)
\((x^2-y^2)(a^2-b^2) = 0\)
Statement 1
\(x^2 > y^2\)
So we can conclude that \((x^2-y^2)\neq{0}\)
Hence \((a^2-b^2) = 0\)
(a+b)(a-b) = 0
We don't know the value of a + b, hence we can't conclude if (a+b) = 0 or (a-b) = 0 or both. Eliminate A and D.
Statement 2
a and b are positive integers.
Insufficient information. We can't find value of a - b. Eliminate B.
Combined
From 1, we know (a+b)(a-b) = 0
From 2, we know that \((a + b)\neq{0}\). This is because a and b are positive, hence the sum must also be +ve.
Therefore (a-b) = 0
Option C
31 May 2024, 12:41
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Hi bunuel, can we not cancel out the squares from both sides initially only ?
