How do you solve the system of equations $9 x - 3 y = 3$ $9x - 3y = 3$ and $3 x + 8 y = - 17$ $3x + 8y = - 17$ ?

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Answer 1 · 2017-03-10T14:55:04

$x = - \frac{1}{3}$ $x=-1/3$

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

Here we have two equations.

$9 x - 3 y = 3 \to$ $9x-3y = 3 ->$ equation $1$ $1$

$3 x + 8 y = - 17 \to$ $3x+8y = -17 ->$ equation $2$ $2$

Multiply equation 2 by $3$ $3$ , we get

$9 x + 24 y = - 51 \to$ $9x + 24y = -51 ->$ equation $3$ $3$

Subtract the new equation $3$ $3$ from $1$ $1$

$9 x - 3 y = 3$ $9x - 3y = 3$

$- (9 x + 24 y = - 51)$ $-(9x +24y = -51)$

$0 x - 27 y = 54$ $0x -27 y = 54$

$\Rightarrow y = - 2$ $=> y=-2$

Substitute in equation $2$ $2$

$3 x + 8 (- 2) = - 17$ $3x + 8(-2) = -17$

$3 x - 16 = - 17$ $3x -16 = -17$

$3 x = - 1$ $3x = -1$

$x = - \frac{1}{3}$ $x= -1/3$

How do you solve the system of equations 9x−3y=39x - 3y = 3 and 3x+8y=−173x + 8y = - 17?