How do solve the following linear system?: $y = 4 x + 4, 3 x - y = - 6$ $y = 4x + 4, 3x - y = -6$ ?

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Answer 1 · 2018-05-21T09:52:24

$x = 2$ $x=2$ and $y = 12$ $y=12$

In two linear equation system, make one variable the subject and substitute that variable in the second equation.

$y = 4 x + 4$ $y = 4x+4$ $\to$ $->$ equation 1
$3 x - y = - 6$ $3x-y=-6$ $\to$ $->$ equation 2

Substitute $y$ $y$ from equation 1 to equation 2:

$3 x - (4 x + 4) = - 6$ $3x - (4x+4) = -6$

$3 x - 4 x - 4 = - 6$ $3x-4x-4 = -6$

$- x - 4 = - 6$ $-x-4 = -6$

$- x = - 6 + 4$ $-x = -6+4$

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

$x = 2$ $x = 2$

Substitute $x = 2$ $x=2$ in equation 1 or equation 2 to get value of $y$ $y$ .
Lets take equation 1:

$y = 4 x + 4$ $y=4x+4$ $\to$ $->$ equation 1

$y = 4 (2) + 4$ $y=4(2) + 4$

$y = 8 + 4$ $y=8+4$

$y = 12$ $y=12$

So we get $x = 2$ $x=2$ and $y = 12$ $y=12$

Check the answer:

$3 x - y = - 6$ $3x-y=-6$
$3 (2) - 12 = - 6$ $3(2) - 12 = -6$
$6 - 12 = - 6$ $6-12=-6$
$- 6 = - 6$ $-6=-6$ $\to$ $->$ Correct

How do solve the following linear system?: y=4x+4,3x−y=−6 y = 4x + 4, 3x - y = -6 ?