Question

Solve the following system of equations by cross-multiplications method.

${\displaystyle a(x+y)+b(x–y)=a^2–ab+b^2}$

${\displaystyle a(x+y)–b(x–y)=a^2+ab+b^2}$

Answer 1

The given system of equations can be rewritten as

${\displaystyle ax+bx+ay–by–(a^2–ab+b^2)=0}$

${\displaystyle \Rightarrow (a+b)x+(a–b)y–(a^2–ab+b^2)=0\dots .\left(1\right)}$

${\displaystyle a(x+y)–b(x–y)=a^2+ab+b^2}$

${\displaystyle \Rightarrow (a–b)x+(a+b)y–(a^2+ab+b^2)=0\dots \left(2\right)}$

Now, by cross-multiplication method, we have

${\displaystyle \Rightarrow \frac{x}{(a-b)\times \{-(a^2+ab+b^2)\}-(a+b)\times \{-(a^2-ab+b^2)\}}=\frac{-y}{(a+b)\times \{-(a^2+ab+b^2)\}-(a-b)\times \{-(a^2-ab+b^2)\}}=\frac{1}{(a+b)\times (a+b)-(a-b)(a-b)}}$

${\displaystyle \Rightarrow \frac{x}{-(a-b)(a^2+ab+b^2)+(a+b)(a^2-ab+b^2)}=\frac{-y}{-(a+b)(a^2+ab+b^2)+(a-b)(a^2-ab+b^2)}=\frac{1}{{(a+b)}^2-{(a-b)}^2}}$

${\displaystyle \Rightarrow \frac{x}{-(a^3-b^3)+(a^3+b^2)}=\frac{-y}{-a^3-2a^{2}b-2a{b}^2-b^3+a^3-2a^{2}b+2a{b}^2-b^3}=\frac{1}{a^2+2ab+b^2-a^2+2ab-b^2}}$

${\displaystyle \Rightarrow \frac{x}{2b^3}=\frac{-y}{-4a^{2}b-2b^3}=\frac{1}{4ab}}$

${\displaystyle \Rightarrow \frac{x}{2b^3}=\frac{-y}{-2b(2a^2+b^2)}=\frac{1}{4ab}}$

${\displaystyle \Rightarrow \frac{x}{2b^3}=\frac{1}{4ab}\Rightarrow x=\frac{b^2}{2a}}$

${\displaystyle \text{and}\frac{-y}{-2b(2a^2+b^2)}=\frac{1}{4ab}\Rightarrow y=\frac{2a^2+b^2}{2a}}$

Hence, the solution is ${\displaystyle x=\frac{b^2}{2a},y=\frac{2a^2+b^2}{2a}}$