How do you solve by substitution $- x + 2 y = - 6$ $-x + 2y = -6$ and $3 x + y = 8$ $3x + y = 8$ ?

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Answer 1 · 2018-04-29T22:44:30

$(\frac{22}{7}, - \frac{10}{7})$ $(22/7,-10/7)$

In the second equation, we can easily solve for $y$ $y$ by subtracting $3 x$ $3x$ from both sides. We get:

$y = 8 - 3 x$ $y=8-3x$

We've already solved for one variable in terms of the other, so we can plug into the first equation. We get

$- x + 2 (8 - 3 x) = - 6$ $-x+2(8-3x)=-6$

$\Rightarrow - x + 16 - 6 x = - 6$ $=>-x+16-6x=-6$

$\Rightarrow - 7 x + 16 = - 6$ $=>-7x+16=-6$

We can subtract $16$ $16$ from both sides to get

$- 7 x = - 22$ $-7x=-22$

Dividing both sides by $- 7$ $-7$ , we get

$x = \frac{22}{7}$ $color(blue)(x=22/7)$

We can plug into our equation for $y$ $y$ , $y = 8 - 3 x$ $y=8-3x$ . We get

$y = 8 - 3 (\frac{22}{7})$ $y=8-3(22/7)$

$y = \frac{56}{7} - \frac{66}{7}$ $y=56/7-66/7$

$y = - \frac{10}{7}$ $color(lime)(y=-10/7)$

Hope this helps!

How do you solve by substitution −x+2y=−6-x + 2y = -6 and 3x+y=83x + y = 8?