Question

Solve the following system of equations.
-3x + 5y = 80
x + 5y = 40
A. x = 8, y = 6.4
B. x = -10, y = 10
C. x = 2, y = 17.2
D. x = 10, y = -10
@jojo4eva @iGreen @chosenmatt

Answer 1

so? what do you think?

Answer 2

A

Answer 3

are u sure?

Answer 4

A is incorrect..

You can solve using substitution..first rearrange the 2nd one:

x + 5y = 40

Subtract 5y to both sides:

x = -5y + 40

Now plug in '-5y + 40' for 'x' in the 1st equation:

-3x + 5y = 80

-3(-5y + 40) + 5y = 80

Distribute -3 into the parenthesis:

15y - 120 + 5y = 80

Combine like terms:

20y - 120 = 80

Add 120 to both sides, what's 120 + 80? @leeaah

Answer 5

200 @iGreen

Answer 6

Yes, that gives us:

20y = 200

Now divide 20 to both sides, what's 200 / 20?

Answer 7

-3x + 5y = 80
x + 5y = 40

You can do this a few ways. but one way would be to turn the +5y in the equation the opposite of each other so they cancel out and then solve for x. To do that what number can we times the equation by to get that? For -3x + 5y = 80 we can times it by -1
-1(-3x + 5y) = 80 *-1
3x + -5y = -80

Now we have
3x - 5y = -80
x + 5y = 40

Cancel out the 5, which will give us
3x = -80
x = 40

Can you finish to solve for x or do you still need help?

Answer 8

Can you divide 200 / 20?

Answer 9

10

Answer 10

Yes, that gives us y = 10. And only one option has that..and what's that?

Answer 11

d

Answer 12

um no

Answer 13

so its b

Answer 14

yeah good job :)

Answer 15

Yes, B is correct.

Answer 16

D has y = -10, not 10. So it's incorrect.