How do you solve $- 3 x + y = - 4$ $-3x+y=-4$ and $2 x + y = 1$ $2x+y=1$ using substitution?

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Answer 1 · 2016-03-04T16:56:56

The solution for the system of equations is:
$x = 1$ $color(green)(x=1$
$y = - 1$ $color(green)(y=-1$

$- 3 x + y = - 4$ $-3x+y=-4$ .......equation $(1)$ $(1)$

$2 x + y = 1$ $2x +y=1$

$y = 1 - 2 x$ $color(green)(y=1-2x$ ..................equation $(2)$ $(2)$

Substituting equation $(2)$ $(2)$ in $(1)$ $(1)$ :

$- 3 x + y = - 4$ $-3x+y=-4$

$- 3 x + 1 - 2 x = - 4$ $-3x+color(green)(1-2x)=-4$

$- 5 x + 1 = - 4$ $-5x +1=-4$

$- 5 x = - 4 - 1$ $-5x =-4-1$

$- 5 x = - 5$ $-5x =-5$

$x = \frac{- 5}{- 5}$ $x =(-5)/(-5)$

$x = 1$ $color(green)(x=1$

Finding $y$ $y$ from equation $2$ $2$ :
$y = 1 - 2 x$ $y=1-2x$

$y = 1 - 2 \times 1$ $y=1-2 xx 1$

$y = 1 - 2$ $y=1-2$

$y = - 1$ $color(green)(y=-1$

How do you solve −3x+y=−4-3x+y=-4 and 2x+y=12x+y=1 using substitution?