How do you solve the system of equations $3 x + 2 y = - 7$ $3x + 2y = - 7$ and $2 x - 7 y = 37$ $2x - 7y = 37$ ?

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Answer 1 · 2018-04-15T05:38:44

$x = 1$ $x=1$
$y = - 5$ $y=-5$

$3 x + 2 y = - 7$ $3x+2y=-7$ --- (1)
$2 x - 7 y = 37$ $2x-7y=37$ --- (2)

From (1),

$3 x + 2 y = - 7$ $3x+2y=-7$
$3 x = - 2 y - 7$ $3x=-2y-7$
$x = \frac{1}{3} (- 2 y - 7)$ $x=1/3(-2y-7)$ --- (3)

Sub (3) into (2),

$2 \times \frac{1}{3} (- 2 y - 7) - 7 y = 37$ $2times1/3(-2y-7)-7y=37$

$\frac{2}{3} (- 2 y - 7) - 7 y = 37$ $2/3(-2y-7)-7y=37$

$- \frac{4}{3} y - \frac{14}{3} - 7 y = 37$ $-4/3y-14/3-7y=37$

$- \frac{25}{3} y = \frac{125}{3}$ $-25/3y=125/3$

$y = - 5$ $y=-5$ --- (4)

Sub (4) into (3),

$x = \frac{1}{3} (- 2 (- 5) - 7)$ $x=1/3(-2(-5)-7)$

$x = \frac{1}{3} (3)$ $x=1/3(3)$

$x = 1$ $x=1$

How do you solve the system of equations 3x+2y=−73x + 2y = - 7 and 2x−7y=372x - 7y = 37?