How do you solve $y = 4 x + 3$ $y = 4x + 3$ and $2 x + 3 y = - 5$ $2x + 3y = -5$ using substitution?

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Answer 1 · 2016-03-21T13:26:25

$x = y = - 1$ $x=y=-1$

$y = 4 x + 3$ $color(blue)(y=4x+3$

$2 x + 3 y = - 5$ $color(blue)(2x+3y=-5$

In the first equation substitute the value of $y$ $y$ to the second equation

$\to 2 x + 3 (4 x + 3) = - 5$ $rarr2x+3(4x+3)=-5$

Use distributive property $a (b + c) = a b + a c$ $color(brown)(a(b+c)=ab+ac$

$\to 2 x + 12 x + 9 = - 5$ $rarr2x+12x+9=-5$

$\to 14 x + 9 = - 5$ $rarr14x+9=-5$

$\to 14 x = - 5 - 9$ $rarr14x=-5-9$

$\to 14 x = - 14$ $rarr14x=-14$

$\Rightarrow x = - \frac{14}{14} = - 1$ $rArrcolor(green)(x=-14/14=-1$

Substitute the value of $x$ $x$ to the first equation

$\to y = 4 (- 1) + 3$ $rarry=4(-1)+3$

$\to y = - 4 + 3$ $rarry=-4+3$

$\Rightarrow y = - 1$ $rArrcolor(green)(y=-1$

$:.color(blue)(ul bar |x=y=-1|$

How do you solve y = 4x + 3y=4x+3y = 4x + 3 and 2x + 3y = -52x+3y=−52x + 3y = -5 using substitution?