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Mathematics

Solve the following pairs of linear equations by cross-multiplication method:

x - y = a + b

ax + by = a2 - b2

Linear Equations

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Answer

Given equations can be written as,

x - y - (a + b) = 0

ax + by - (a2 - b2) = 0

By cross multiplication method,

Solve by cross-multiplication x - y = a + b and ax + by = a^2 - b^2. Simultaneous Linear Equations, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

∴x(−1)×[−(a2−b2)]−b×[−(a+b)]=y[−(a+b)×a]−[−(a2−b2)×1]=11×b−a×(−1)⇒xa2−b2+ab+b2=y−a2−ab+a2−b2=1b+a⇒xa2+ab=y−ab−b2=1b+a∴xa2+ab=1b+a and y−b(a+b)=1b+a⇒xa(a+b)=1b+a and y−b(a+b)=1b+a⇒x=a(a+b)b+a and y=−b(a+b)b+a⇒x=a and y=−b.\therefore \dfrac{x}{(-1) \times [-(a^2 - b^2)] - b \times [-(a + b)]} = \dfrac{y}{[-(a + b) \times a] - [-(a^2 - b^2) \times 1] } = \dfrac{1}{1 \times b - a \times (-1)} \\[1em] \Rightarrow \dfrac{x}{a^2 - b^2 + ab + b^2} = \dfrac{y}{-a^2 - ab + a^2 - b^2} = \dfrac{1}{b + a} \\[1em] \Rightarrow \dfrac{x}{a^2 + ab} = \dfrac{y}{-ab - b^2} = \dfrac{1}{b + a} \\[1em] \therefore \dfrac{x}{a^2 + ab} = \dfrac{1}{b + a} \text{ and } \dfrac{y}{-b(a + b)} = \dfrac{1}{b + a} \\[1em] \Rightarrow \dfrac{x}{a(a + b)} = \dfrac{1}{b + a} \text{ and } \dfrac{y}{-b(a + b)} = \dfrac{1}{b + a} \\[1em] \Rightarrow x = \dfrac{a(a + b)}{b + a} \text{ and } y = \dfrac{-b(a + b)}{b + a} \\[1em] \Rightarrow x = a \text{ and } y = -b.

Hence, x = a and y = -b.

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