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If x(a + b) + y(a - b) = 2 and ax + by = (a^2 + b^2)/(a^2 - b^2), then express 1/x + 1/y in terms of a and b.

 May 19, 2020
 #1
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If
\(x(a + b) + y(a - b) = 2\)
and
\(ax + by = \dfrac{a^2 + b^2} {a^2 - b^2}\),
then express \(\dfrac{1}{x} + \dfrac{1}{y}\) in terms of \(a\) and \(b\).

 

\(\begin{array}{|rcll|} \hline \mathbf{x(a + b) + y(a - b)} &=& \mathbf{2} \\\\ y(a - b) &=& 2-x(a + b) \\\\ \mathbf{y} &=& \mathbf{\dfrac{2-x(a + b)} {a - b}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{ax + by } &=& \mathbf{ \dfrac{a^2 + b^2} {a^2 - b^2}} \\\\ ax + b\left( \dfrac{2-x(a + b)} {a - b} \right) &=& \dfrac{a^2 + b^2} {a^2 - b^2} \\\\ ax + b\left( \dfrac{2-x(a + b)} {a - b} \right) &=& \dfrac{a^2 + b^2} {(a-b)(a+b)} \quad | \quad *(a-b) \\\\ a(a-b)x + b\Big(2-x(a + b) \Big) &=& \dfrac{a^2 + b^2} {a+b} \\\\ a(a-b)x + 2b-xb(a + b) &=& \dfrac{a^2 + b^2} {a+b} \\\\ x\Big( a(a-b)-b(a + b)\Big) + 2b &=& \dfrac{a^2 + b^2} {a+b} \\\\ x( a^2-2ab-b^2) + 2b &=& \dfrac{a^2 + b^2} {a+b} \\\\ x( a^2-2ab-b^2) &=& \dfrac{a^2 + b^2} {a+b} -2b \\\\ x( a^2-2ab-b^2) &=& \dfrac{a^2 + b^2-2b(a+b)} {a+b} \\\\ x( a^2-2ab-b^2) &=& \dfrac{a^2 + b^2-2ab-2b^2} {a+b} \\\\ x( a^2-2ab-b^2) &=& \dfrac{a^2 -2ab -b^2} {a+b} \quad | \quad :a^2-2ab-b^2 \\\\ x&=& \dfrac{1} {a+b} \\\\ \mathbf{\dfrac{1}{x}} &=& \mathbf{a+b} \\ \hline \mathbf{y} &=& \mathbf{\dfrac{2-x(a + b)} {a - b}} \\\\ \dfrac{1}{y} &=& \dfrac {a - b} {2-x(a + b)} \quad | \quad \mathbf{x=\dfrac{1} {a+b}} \\\\ \dfrac{1}{y} &=& \dfrac {a - b} {2-\dfrac{1} {(a+b)}(a + b)} \\\\ \dfrac{1}{y} &=& \dfrac {a - b} {2-1} \\\\ \mathbf{\dfrac{1}{y}} &=& \mathbf{a-b} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \dfrac{1}{x} + \dfrac{1}{y} &=& (a+b) + (a-b) \\\\ \mathbf{\dfrac{1}{x} + \dfrac{1}{y}} &=& \mathbf{2a} \\ \hline \end{array}\)

 

laugh

 May 20, 2020

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