How do you solve the following system: $\frac{1}{2} x - 2 y = 4, 5 x + 3 y = - 1$ $1/2x - 2y = 4 , 5x+3y=-1$ ?

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How do you solve the following system: $\frac{1}{2} x - 2 y = 4, 5 x + 3 y = - 1$ $1/2x - 2y = 4 , 5x+3y=-1$ ?

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Answer 1 · 2017-05-07T15:25:46

$x = \frac{20}{23} and y = - \frac{41}{23}$ $x = 20/23 and y = -41/23$

Explanation:

The approach is either to use substitution or elimination.

For substitution, you need to have a single variable such as $x$ $x$

$\frac{1}{2} x - 2 y = 4 and 5 x + 3 y = - 1$ $1/2x-2y =4" and "5x+3y =-1$

Multiply the first equation by $2$ $2$

$2 \times \frac{1}{2} x - 2 \times 2 y = 2 \times 4$ $2xx1/2x -2xx2y =2xx 4$

$x - 4 y = 8$ $x-4y =8$

$x = (4 y + 8)$ $color(blue)(x = (4y+8))" "$ Substitute for $x$ $x$ in the other equation:

$m l . m 5 x + 3 y = - 1$ $color(white)(ml.m)5color(blue)(x)+3y=-1$
$m m m ↓$ $color(white)(mmm)darr$
$m 5 (4 y + 8) + 3 y = - 1$ $color(white)(m)5color(blue)((4y+8)) +3y=-1$

$20 y + 40 + 3 y = - 1$ $20y+40 +3y =-1$

$23 y = - 1 - 40$ $23y = -1-40$

$23 y = - 41$ $23y = -41$

$y = \frac{- 41}{23}$ $y = (-41)/23$

Now find $x by using x = 4 y + 8$ $x" by using " x = 4y+8$

$x = 4 (\frac{- 41}{23}) + 8$ $x = 4((-41)/23)+8$

$x = \frac{20}{23}$ $x = 20/23$

Substitute to check:

$5 x + 3 y$ $5x +3y$

$5 (\frac{20}{23}) + 3 (- \frac{41}{23})$ $5(20/23)+3(-41 /23)$

$= \frac{100}{23} - \frac{123}{23}$ $=100/23-123/23$

$= - 1$ $=-1$

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How do you solve the following system: $\frac{1}{2} x - 2 y = 4, 5 x + 3 y = - 1$ $1/2x - 2y = 4 , 5x+3y=-1$ ?

1 Answer

Explanation:

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How do you solve the following system: $\frac{1}{2} x - 2 y = 4, 5 x + 3 y = - 1$ $1/2x - 2y = 4 , 5x+3y=-1$ ?