How do solve the following linear system?: $2 x + 4 = - 4 y, 7 x + 6 y + 11 = 0$ $2x + 4 = -4y, 7x + 6y + 11 = 0$ ?

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Answer 1 · 2017-06-17T17:16:22

$(- \frac{5}{4}, - \frac{3}{8})$ $(-5/4,-3/8)$

First, solve one of the equations for $x$ $x$ . It doesn't matter which one.

$2 x + 4 = - 4 y$ $2x+4 = -4y$

$2 x = - 4 - 4 y$ $2x=-4-4y$

$x = - 2 - 2 y$ $x=-2-2y$

Next, plug this equation for $x$ $x$ into the other equation

$7 x + 6 y + 11 = 0$ $7color(red)(x)+6y+11=0$

$7 (- 2 - 2 y) + 6 y + 11 = 0$ $7(color(red)(-2-2y))+6y+11=0$

$- 14 - 14 y + 6 y + 11 = 0$ $-14-14y+6y+11=0$

$- 14 - 8 y + 11 = 0$ $-14-8y+11=0$

$- 3 - 8 y = 0$ $-3-8y=0$

$- 8 y = 3$ $-8y=3$

$y = - \frac{3}{8}$ $y=-3/8$

Finally, plug this solution for $y$ $y$ into the equation for $x$ $x$

$x = - 2 - 2 y$ $x=-2-2y$

$x = - 2 - 2 (- \frac{3}{8})$ $x=-2-2(-3/8)$

$x = - 2 + \frac{6}{8}$ $x=-2+6/8$

$x = - \frac{16}{8} + \frac{6}{8} = - \frac{10}{8} = - \frac{5}{4}$ $x=-16/8+6/8=-10/8=-5/4$

How do solve the following linear system?: 2x + 4 = -4y, 7x + 6y + 11 = 0 2x+4=−4y,7x+6y+11=02x + 4 = -4y, 7x + 6y + 11 = 0 ?