Solve the following equations by substitution method.
$3 a - 2 b = - 10; 2 a + 3 b = 2$

a = - 2, b = 2

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a = - 1, b = 2

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a = - 4, b = 2

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a = - 7, b = 2

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Solution

The correct option is A $a = - 2, b = 2$
$3 a - 2 b = - 10; 2 a + 3 b = 2$
$a = \frac{2 b - 10}{3} \Rightarrow$ Put it in second equation.
$2 \times \frac{2 b - 10}{3} + 3 b = 2$
$4 b - 20 + 9 b = 6$
$13 b = 26$
$∴ b = 2$

$a = \frac{2 b - 10}{3} = \frac{2 (2) - 10}{3} = - 2$
$a = - 2$ and $b = 2$

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Similar questions

(a + b) x + (a - b) y = 3 a - b (2 a^{2} + b^{2}) x - (a^{2} - b) y = b^{2} + 2 b

Solve the equation by substitution method :

(a+b)x + (a+b)y = 2a² -2b²

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

a^{2} b^{2} x^{2} - (a^{2} + b^{2}) x + 1 = 0

\frac{2 x}{a} + \frac{y}{b} = 2

\frac{x}{a} - \frac{y}{b} = 4

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