How do you solve the system $- 2 x + y = 8 and y = - 3 x - 2$ $-2x+y=8 and y=-3x-2$ ?

Question

How do you solve the system $- 2 x + y = 8 and y = - 3 x - 2$ $-2x+y=8 and y=-3x-2$ ?

2 Answers

Impact of this question

3648 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-09T18:46:36

$x = - 2, y = 4$ $x=-2,y=4$

Explanation:

Plugging the second equation

$y = - 3 x - 2$ $y=-3x-2$ in the first one:

$- 3 x - 2 x - 2 = 8$ $-3x-2x-2=8$
adding $2$ $2$ on both sides and collexting like Terms

$- 5 x = 10$ $-5x=10$

$x = - 2$ $x=-2$ so $y = 6 - 2 = 4$ $y=6-2=4$

Answer 2 · 2018-07-09T18:54:12

$(x, y) \to (- 2, 4)$ $(x,y)to(-2,4)$

Explanation:

$- 2 x + y = 8 \to (1)$ $-2x+y=8to(1)$

$y = - 3 x - 2 \to (2)$ $y=-3x-2to(2)$

$equation (2) expresses y in terms of x$ $"equation "(2)" expresses y in terms of "x$

$substitute y = - 3 x - 2 into equation (1)$ $"substitute "y=-3x-2" into equation "(1)$

$- 2 x - 3 x - 2 = 8$ $-2x-3x-2=8$

$- 5 x - 2 = 8$ $-5x-2=8$

$add 2 to both sides$ $"add 2 to both sides"$

$- 5 x = 8 + 2 = 10$ $-5x=8+2=10$

$divide both sides by - 5$ $"divide both sides by "-5$

$x = \frac{10}{- 5} \Rightarrow x = - 2$ $x=10/(-5)rArrx=-2$

$substitute x = - 2 into equation (2)$ $"substitute "x=-2" into equation "(2)$

$y = 6 - 2 = 4$ $y=6-2=4$

$the point of intersection = (- 2, 4)$ $"the point of intersection "=(-2,4)$
graph{(y+3x+2)(y-2x-8)=0 [-10, 10, -5, 5]}

Science

Math

Humanities

... and beyond

How do you solve the system $- 2 x + y = 8 and y = - 3 x - 2$ $-2x+y=8 and y=-3x-2$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve the system -2x+y=8 and y=-3x-2−2x+y=8andy=−3x−2-2x+y=8 and y=-3x-2?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you solve the system $- 2 x + y = 8 and y = - 3 x - 2$ $-2x+y=8 and y=-3x-2$ ?