# Solve the Following System of Equations by Cross-multiplications Method - Mathematics

Sum

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

a(x + y) + b (x – y) = a^2 – ab + b^2

a(x + y) – b (x – y) = a^2 + ab + b^2

#### Solution

The given system of equations can be rewritten as

ax + bx + ay – by – ( a^2 – ab + b^2 ) = 0

⇒ (a + b) x + (a – b) y – (a^2 – ab + b^2 ) = 0 ….(1)

a(x + y) – b (x – y) = a^2 + ab + b^2

⇒ (a – b) x + (a + b) y – (a^2 + ab + b^2 ) = 0 …(2)

Now, by cross-multiplication method, we have

\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)}

\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}

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

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

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

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

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

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

Concept: Algebraic Methods of Solving a Pair of Linear Equations - Cross - Multiplication Method
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