How do you solve the following system?: $3x +2y =8 , 6x +9y = -3$

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Answer 1 · 2015-12-01T00:04:37

$x = 26/5$ and $y = -19/5$

We multiply the first equation by $2$ to get the same coefficient for $x$ in both equations. The system then becomes

$6x + 4y = 16$

$6x + 9y = -3$

Subtract the first equation from the second:

$6x + 9y = -3$

$- (6y + 4y = 16)$

In other words we add vertically:

$6x + 9y = -3$

$- 6y - 4y = -16$

The result is

$0 + 5y = -19$

So $5y = -19$ and dividing both sides by $5$ we get $y = -19/5$ .
We substitute this value of y into the first equation:

$3x - 38/5 = 8$

Therefore $3x = 8 + 38/5$ (subtracting $2y = -38/5$ from each side),
So

$3x = (40 + 38)/5 = 78/5$ and $x = 26/5$

How do you solve the following system?: 3x +2y =8 , 6x +9y = -33x +2y =8 , 6x +9y = -3