Question #8c1e4

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Answer 1 · 2017-02-05T06:50:55

$x = 3$ $x=3$ and $y = 1$ $y=1$

$- 3 x + 2 y = - 7$ $-3x+2y=-7$
$3 x + 3 y = 12$ $3x+3y=12$

From the first equation, derive a value for $3 x$ $3x$ .

$- 3 x + 2 y = - 7$ $-3x+2y=-7$

Multiply all terms with $- 1$ $-1$ .

$3 x - 2 y = 7$ $3x-2y=7$

Add $2 y$ $2y$ to both sides.

$3 x = 7 + 2 y$ $3x=7+2y$

In the second equation, substitute $3 x$ $3x$ with $(7 + 2 y)$ $color(red)((7+2y))$ .

$3 x + 3 y = 12$ $3x+3y=12$

$(7 + 2 y) + 3 y = 12$ $color(red)((7+2y))+3y=12$

Open the brackets and simplify.

$7 + 2 y + 3 y = 12$ $color(red)(7+2y)+3y=12$

$7 + 5 y = 12$ $7+5y=12$

Subtract $7$ $7$ from each side.

$5 y = 5$ $5y=5$

Divide both sides by $5$ $5$ .

$y = 1$ $y=1$

In the second equation, substitute $y$ $y$ with $1$ $color(blue)(1)$ .

$3 x + 3 y = 12$ $3x+3y=12$

$3 x + 3 (1) = 12$ $3x+3color(blue)((1))=12$

Open the brackets and simplify.

$3 x + 3 = 12$ $3x+3=12$

Subtract $3$ $3$ from each side.

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

Divide both sides by $3$ $3$ .

$x = 3$ $x=3$