Solve the following equations by substitution method.
$2 x - 3 y = 14; 5 x + 2 y = 16$

x = 7, y = 1

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x = 4, y = - 2

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x = 6, y = - 2

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x = 3, y = - 1

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Solution

The correct option is B $x = 4, y = - 2$
$2 x - 3 y = 14; 5 x + 2 y = 16$
$x = \frac{14 + 3 y}{2}$ Put it in second equation.
$5 \times \frac{14 + 3 y}{2} + 2 y = 16$
$70 + 15 y + 4 y = 32$
$19 y = - 38$
$y = - 2$

$x = \frac{14 + 3 y}{2} = \frac{14 + 3 (- 2)}{2} = 4$
$x = 4$ and $y = - 2$

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Similar questions

x = 2 y - 1; y = 2 x - 7

Solve the following pair of linear equations by the elimination method and the substitution method :

(i) x+y=5 and 2x−3y=4
(ii) 3x+4y=10 and 2x−2y=2
(iii) 3x−5y−4=0 and 9x=2y+7
(iv) x2+2y3=−1 and x−y3=3

\frac{3}{x} + \frac{1}{y} = 7; \frac{5}{x} - \frac{4}{y} = 6. (x \neq 0, y \neq 0)

x - y = 1

2 x - 3 y = 2

\frac{1}{2 x} + \frac{1}{3 y} = 2 (x \neq 0, y \neq 0) \frac{1}{3 x} + \frac{1}{y} = \frac{11}{3}

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