Solve the following equations by substitution method.
$3 y - 2 x = 9; 2 x + 5 y = 15$

x = 0, y = 2

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x = 5, y = 2

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x = 1, y = 3

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x = 0, y = 3

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Solution

The correct option is B $x = 0, y = 3$
$3 y - 2 x = 9; 2 x + 5 y = 15$
$y = \frac{9 + 2 x}{3}$ Put it in second equation.
$2 x + 5 \times \frac{9 + 2 x}{3} = 15$
$6 x + 45 + 10 x = 45$
$16 x = 0$
$x = 0$
$y = \frac{9 + 2 x}{3} = \frac{9 + 2 (0)}{3} = 3 y = 3$

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