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

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Answer 1 · 2016-05-18T17:11:07

The solution for the system of equations is:

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

$y = 2 x - 3$ $color(blue)(y = 2x - 3$ ............equation $(1)$ $(1)$

$y = - 3 x + 2$ $y = -3x + 2$ ......equation $(2)$ $(2)$

Solving by substitution.

Substituting equation $1$ $1$ in equation $2$ $2$ :

$y = - 3 x + 2$ $y = -3x + 2$

$2 x - 3 = - 3 x + 2$ $color(blue)( 2x - 3)= -3x + 2$

$2 x + 3 x = 2 + 3$ $2x + 3x= 2 + 3$

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

$x = \frac{5}{5}$ $x = 5/5$

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

Finding the value of $y$ $y$ from equation $1 :$ $1:$

$y = 2 x - 3$ $y = 2x - 3$

$y = (2 \cdot 1) - 3$ $y = (2 * 1) - 3$

$y = 2 - 3$ $y = 2 - 3$

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

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