How do you solve $3 y = - 2 x, y = x - 5$ $3y=-2x, y=x-5$ by substitution?

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How do you solve $3 y = - 2 x, y = x - 5$ $3y=-2x, y=x-5$ by substitution?

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Answer 1 · 2016-12-06T23:42:33

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

Explanation:

Step 1) Because the second equation is already solved for $y$ $y$ we can substitute $x - 5$ $x - 5$ for $y$ $y$ in the first equation and solve for $x$ $x$ :

$3 (x - 5) = - 2 x$ $3(x - 5) = -2x$

$3 x - 15 = - 2 x$ $3x - 15 = -2x$

$3 x - 15 + 15 + 2 x = - 2 x + 15 + 2 x$ $3x - 15 + 15 + 2x = -2x + 15 + 2x$

$3 x - 0 + 2 x = 0 + 15$ $3x - 0 + 2x = 0 +15$

$5 x = 15$ $5x = 15$

$\frac{5 x}{5} = \frac{15}{5}$ $(5x)/5 = 15/5$

$x = 3$ $x = 3$

Step 2) Substitute $3$ $3$ for $x$ $x$ in the second equation and calculate $y$ $y$ :

$y = 3 - 5$ $y = 3 - 5$

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

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How do you solve $3 y = - 2 x, y = x - 5$ $3y=-2x, y=x-5$ by substitution?

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How do you solve $3 y = - 2 x, y = x - 5$ $3y=-2x, y=x-5$ by substitution?