Question

please help

Answer 1

@iGreen

Answer 2

@ChantySquirrel1129**

Answer 3

@seb.cal

Answer 4

@Vocaloid

Answer 5

ok, hold on reaally quick,.

Answer 6

@LegendarySadist

Answer 7

do u know this stuff @LegendarySadist

Answer 8

Yes, I do know it. Do you have any method you want to solve it with? Substitution?

Answer 9

no

Answer 10

Well then let's just solve it with substitution. First, get x alone in the bottom equation.

Answer 11

so do i add 3 to y or y to 3

Answer 12

Y to 3.

Answer 13

so x=3y?

Answer 14

The order doesn't matter. But it should look like \[\large \sf x=3+y\]

Answer 15

ok

Answer 16

so the first one is 4x=2+y?

Answer 17

So now that you know what x is, substitute "3+y" into the first equation. And no, we are only getting x alone for the second equation.

Answer 18

im confused

Answer 19

the second equation is done?

Answer 20

No, we'll get back to it.

Answer 21

ok

Answer 22

so with those substitutions how should the first equation look?

Answer 23

So like I said, for the equation \[\large \sf 4x+y=2\] you will substitute\[\large \sf y+3\] in for every x.

Answer 24

Sooo, the first equation should look like \[\large \sf 4(y+3)+y=2\] and then solve for y

Answer 25

k im confused do i minus y from 2

Answer 26

Well we are solving for y. So we want all of the y values on the left side and all the "regular" numbers, on the right. So first we would distribute the 4 onto the y and the 3. Then we would move "regular" numbers to the right side and combine like terms on both sides.

Answer 27

4y+12+y=2 5y=-10 y=-2 right?

Answer 28

Correct. Now we will substitute "y=-2" into either equation and solve for x.

Answer 29

k thanks

Answer 30

No problem :)