Question

If you were to use the substitution method to solve the following system, choose the new equation after the expression equivalent to x from the first equation is substituted into the second equation.
-x + 5y = 1
2x + 4y = -4
Choose one answer.
a. 2(5y - 1) + 4y = -4
b. 2(-5y + 1) + 4y = -4
c. 2x + 4(-5y + 1) = -4
d. 2x + 4(5y - 1) = -4

Answer 1

XD i actually just learned this lol

Answer 2

so you will be putting the first equation into the second. in this one you are solving for x for the first equation, do you know how to do this

Answer 3

can you just show your work, for the whole thing

Answer 4

so the first equation is
-x +5y = 1
You want to solve for x, what that means is getting x alone on one side of the equation.
you can do this many ways, i will add x to each side to get
-x + x + 5y = 1 +x
the x's on the left side cancel to zero
5y = 1 + x
subtract 1 from each side
5y - 1 = 1 - 1 + x
5y - 1 = x :since 1 - 1 = 0

now that you have x = 5y - 1 you can put it into the second equation

2x + 4y = -4 :just replace x with 5y -1 since you know they are equal they are the same thing

2( 5y-1) + 4y = -4

Answer 5

can you help me with another

Answer 6

sure thing

Answer 7

If you were to use the substitution method to solve the following system, choose the new system of equations that would result if z was isolated in the first equation.

2x - 5y + z = 15
4x - y + 3z = 15
3x - 2y + 2z = 13

Answer 8

this time i want you to get z on its own,
2x - 5y + z = 15
you want z on one side and the rest on the other. Once you get that i will continue

Answer 9

alright one sec

Answer 10

is it z=15+5y

Answer 11

z=15+5y-2x

Answer 12

yes it is.

Answer 13

Now that you have z, z=15+5y-2x, you just replace that z into both of the equations
second equation:
4x - y + 3z = 15

so putting z in:
4x - y + 3(15+5y-2x) = 15
third equation:
3x - 2y + 2z = 13

with z substituted:
3x - 2y + 2(15+5y-2x) = 13

Answer 14

Now that more difficult part is to simplify both of these all down to just one x term, one y term, and one constant each. But once you do it would be just like the first question with 2 variables

Answer 15

now the* more difficult part

Answer 16

do you need to see how the simplification would go?

Answer 17

no it's ok

Answer 18

i follow you