How do you solve the system $7 x + 2 y = 2$ $7x + 2y = 2$ and $6 x - 2 y = 8$ $6x - 2y = 8$ ?

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Answer 1 · 2015-05-16T16:44:05

I notice that all the terms in the second equation are divisible by 2 and it will be easy to isolate $y$ $y$ , so let's do that first.

$6 x - 2 y = 8$ $6x-2y=8$

Divide both sides by 2 to get

$3 x - y = 4$ $3x-y=4$

Add $y$ $y$ to both sides to get

$3 x = 4 + y$ $3x=4+y$

Subtract 8 from both sides to get

$y = 3 x - 4$ $y = 3x-4$

Now substitute this expression for $y$ $y$ into the first equation...

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

$= 7 x + 2 (3 x - 4)$ $= 7x + 2(3x-4)$

$= 7 x + 6 x - 8$ $= 7x + 6x - 8$

$= 13 x - 8$ $=13x - 8$

Add 8 to both sides to get

$13 x = 10$ $13x=10$

Divide both sides by 13 to get

$x = \frac{10}{13}$ $x = 10/13$

$y = 3 x - 4 = \frac{3 \times 10}{13} - 4$ $y = 3x-4 = (3xx10)/13-4$

$= \frac{30}{13} - \frac{4 \times 13}{13}$ $= 30/13-(4xx13)/13$

$= \frac{30 - 52}{13}$ $= (30-52)/13$

$= - \frac{22}{13}$ $= -22/13$

How do you solve the system 7x+2y=27x + 2y = 2 and 6x−2y=86x - 2y = 8?