# Solve the following system of equations by the method of cross-multiplication x/a=y/b=a+b; (x/a^2)+(y/b^2)=2 - Mathematics

Sum

Solve the following system of equations by the method of cross-multiplication \frac{x}{a}+\frac{y}{b}=a+b ;   \frac{x}{a^{2}}+\frac{y}{b^{2}}=2

#### Solution

The given system of equations is rewritten as:

\frac{x}{a}+\frac{y}{b}-( a+b) ….(1)

\frac{x}{a^{2}}+\frac{y}{b^{2}}-2 ….(2)

Multiplying equation (1) by ab, we get

bx + ay – ab (a + b) = 0 ….(3)

Multiplying equation (2) by a^2 b^2 , we get

b^2 x + a^2 y – 2a^2 b^2 = 0 ….(4)

By cross multiplication method, we have

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

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

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

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

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

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

And\text{ }\frac{y}{ab^{3}(a-b)}=\frac{1}{ab(a-b)}

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

Hence, the solution x = a2 , y = b2

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