How do you solve the following system: $4 x + y = - 1, 5 x - 7 y = 12$ $4x + y = -1 , 5x - 7y = 12$ ?

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How do you solve the following system: $4 x + y = - 1, 5 x - 7 y = 12$ $4x + y = -1 , 5x - 7y = 12$ ?

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Answer 1 · 2018-04-13T15:25:18

$x = \frac{5}{33}$ $x=5/33$
$y = - \frac{5}{3}$ $y=-5/3$
$(\frac{5}{33}, - \frac{5}{3})$ $(5/33, -5/3)$

Explanation:

$4 x + y = - 1$ $4x+y=-1$ , $5 x - 7 y = 12$ $5x-7y=12$
Take the first equation, $4 x + y = - 1$ $4x+y=-1$ , and isolate $y$ $y$ :
$y = - 4 x - 1$ $y=-4x-1$ . We can substitute this value into the second equation:
$5 x - 7 (- 4 x - 1) = 12$ $5x-7(-4x-1)=12$ . Distribute the parentheses:
$5 x + 28 x + 7 = 12$ $5x+28x+7=12$ Combine like terms, isolate $x$ $x$ and coeffecient:
$33 x = 5$ $33x=5$ Isolate $x$ $x$ :
$x = \frac{5}{33}$ $x=5/33$ Input this value into the first equation:
$4 (\frac{5}{33}) + y = - 1$ $4(5/33)+y=-1$ Distribute $y$ $y$
$\frac{20}{33} + y = - 1$ $20/33+y=-1$ Isolate $y$ $y$
$y = - \frac{55}{33}$ $y=-55/33$
$y = - \frac{5}{3}$ $y=-5/3$

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How do you solve the following system: $4 x + y = - 1, 5 x - 7 y = 12$ $4x + y = -1 , 5x - 7y = 12$ ?

1 Answer

Explanation:

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How do you solve the following system: $4 x + y = - 1, 5 x - 7 y = 12$ $4x + y = -1 , 5x - 7y = 12$ ?