Solve the following systems of equations by the method of cross-multiplication:
$(a - b) x + (a + b) y = 2 a^{2} - 2 b^{2}$
$(a + b) (x + y) = 4 a b$

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Solution

$(a - b) x + (a + b) y = 2 a^{2} - 2 b^{2} ⟶ (i) (a + b) x + (a + b) y = 4 a b ⟶ (i i)$
On simplifying $,$
$(a - b) x + (a + b) y - (2 a^{2} - 2 b^{2}) = 0$
$(a + b) x + (a + b) y - 4 a b = 0$
On cross multiplication $,$
$\frac{x}{∣ ∣ ∣ \begin{matrix} (a + b) & - (2 a^{2} - 2 b^{2}) (a + b) & - 4 a b \end{matrix} ∣ ∣ ∣} = \frac{- y}{∣ ∣ ∣ \begin{matrix} (a - b) & - (2 a^{2} - 2 b^{2}) (a + b) & - 4 a b \end{matrix} ∣ ∣ ∣} = \frac{1}{∣ ∣ ∣ \begin{matrix} (a - b) & (a + b) (a + b) & (a + b) \end{matrix} ∣ ∣ ∣} \Rightarrow \frac{x}{- 4 a b (a + b) + (2 a^{2} - 2 b^{2}) (a + b)} = \frac{- y}{(a - b) (- 4 a b) + (2 a^{2} - 2 b^{2}) (a + b)} = \frac{1}{a^{2} - b^{2} - {(a + b)}^{2}} ∴ x = \frac{(a + b) (2 a^{2} - 2 b^{2} - 4 a b)}{- 2 b (a + b)} \Rightarrow x = \frac{- a^{2} + b^{2} + 2 a b}{b} - y = \frac{(a - b) (- 4 a b + 2 {(a + b)}^{2})}{- 2 b (a + b)} \Rightarrow y = \frac{(a - b) (a^{2} + b^{2})}{b (a + b)}$

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Similar questions

(a - b) x + (a + b) y = 2 a^{2} - 2 b^{2} (a + b) (x + y) = 4 a b

(a - b) x + (a + b) y = 2 (a^{2} - b^{2})

(a + b) x - (a - b) y = 4 a b

(a - b) x + (a + b) y = 2 a^{2} - 2 a b^{2}, (a + b) (x + y) = 4 a b

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

Solve the equation by substitution method :

(a+b)x + (a+b)y = 2a² -2b²

(a+b) (x + y) = 4ab

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