How do you solve the following system?: $y = 2 x + 3, y = - x + 12$ $y=2x + 3 , y=-x + 12$

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Answer 1 · 2016-04-18T16:50:59

$x = 3$ $color(blue)(x=3$
$y = 9$ $color(blue)(y =9$

$y = 2 x + 3$ $y = 2x + 3$ .............equation $(1)$ $(1)$

$y = - x + 12$ $color(blue)(y = - x +12$ .........equation $(2)$ $(2)$

Solving by substitution.
Substituting equation $2$ $2$ in $1$ $1$ :

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

$- x + 12 = 2 x + 3$ $color(blue)(- x +12)= 2x + 3$

$12 - 3 = 2 x + x$ $12 - 3= 2x + x$

$9 = 3 x$ $9= 3x$

$x = \frac{9}{3}$ $x = 9/3$

$x = 3$ $color(blue)(x=3$

Finding $y$ $y$ from equation $1$ $1$ :
$y = 2 x + 3$ $y = 2x + 3$

$y = 2 \cdot (3) + 3$ $y = 2 * color(blue)((3)) + 3$

$y = 6 + 3$ $y =6 + 3$

$y = 9$ $color(blue)(y =9$

How do you solve the following system?: y=2x+3,y=−x+12y=2x + 3 , y=-x + 12