How do you solve $y = 3 x + 2$ $y=3x+2$ and $y = \frac{4}{3} x - 5$ $y=4/3x-5$ using substitution?

Question

1111 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-23T07:19:50

$x = - 4 \frac{1}{5}$ $x=-4 1/5$ and $y = - 10 \frac{3}{5}$ $y=-10 3/5$

$y = 3 x + 2$ $y=3x+2$
$y = \frac{4}{3} x - 5$ $y=4/3x-5$

In the second equation, substitute $y$ $y$ with $(3 x + 2)$ $color(red)((3x+2))$ . This equivalence is taken from the first equation.

$3 x + 2 = \frac{4}{3} x - 5$ $color(red)(3x+2)=4/3x-5$

Multiply all terms by $3$ $3$ .

$9 x + 6 = 4 x - 15$ $9x+6=4x-15$

Subtract $4 x$ $4x$ from each side.

$5 x + 6 = - 15$ $5x+6=-15$

Subtract $6$ $6$ from each side.

$5 x = - 21$ $5x=-21$

Divide both sides by $5$ $5$ .

$x = - \frac{21}{5}$ $x=-21/5$ or $x = - 4 \frac{1}{5}$ $x=-4 1/5$

In the first equation, substitute $x$ $x$ with $- \frac{21}{5}$ $color(blue)(-21/5)$ .

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

$y = 3 (- \frac{21}{5}) + 2$ $y=3color(blue)((-21/5))+2$

$y = - \frac{63}{5} + 2$ $y=-63/5+2$

$y = \frac{- 63 + 10}{5}$ $y=(-63+10)/5$

$y = - \frac{53}{5}$ $y=-53/5$ or $y = - 10 \frac{3}{5}$ $y=-10 3/5$

How do you solve y=3x+2y=3x+2y=3x+2 and y=4/3x-5y=43x−5y=4/3x-5 using substitution?