Question

# Solve by cross-multiplication method :$$(a - b) x + (a + b) y = 2 (a^2- b^2)$$,$$(a + b)x - (a -b) y = 4ab$$

A
x=(ab) and y=(2ab)
B
x=(a+b) and y=(ab)
C
x=(2a+5b) and y=(ab2)
D
x=(2a3b) and y=(3ab)

Solution

## The correct option is B $$x=(a+b)$$ and $$y=(a-b)$$Writing the equations in the standard form, we get.$$(a - b) x + (a + b) y = 2 (a^2- b^2)$$,$$(a + b)x - (a -b) y = 4ab$$Applying the cross-multiplication method, we get$$- 4ab (a + b) - 2 (a - b) (a^2 - b2)$$$$=- 2(a + b) [2ab + (a - b)^2]$$$$=- 2(a + b) (2ab + a^2 + b^2 - 2ab)$$$$=-2(a + b) (a^2 + b^2)$$Simplification of the expression under $$y$$ :$$- 2(a^2 - b^2) (a + b)+4ab (a - b)$$$$=- 2(a - b) [(a + b) (a + b) - 2ab]$$$$=- 2(a - b) (a^2 + b^2 + 2ab - 2ab)$$$$=-2(a - b) (a^2 + b^2)$$Simplification of the expression under $$1$$ :$$-(a-b)^2-(a+b)^2$$$$=-(a^2+b^2-2ab)-(a^2+b^2+2ab)$$$$=-2(a^2+b^2)$$Hence,$$\dfrac {x}{-2(a+b)(a^2+b^2)}=\dfrac {y}{-2(a-b)(a^2+b^2)}=\dfrac {1}{-2(a^2+b^2)}$$$$\Rightarrow \dfrac {x}{a+b}=\dfrac {y}{a-b}=\dfrac {1}{1}$$$$\Rightarrow x=(a+b)$$ and $$y=(a-b)$$Mathematics

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