How do you work x=2y-2 and 3y=x+6 as substitution?

3 Answers
Apr 13, 2018

=>x = 6x=6
=>y = 4y=4

Explanation:

Substitution involves putting a known variable expression into another expression.

In this case, you are given x = 2y-2x=2y2. You can substitute this into the second equation as follows:

=>3y = x+63y=x+6

=>3y = (2y-2) + 63y=(2y2)+6

=>3y = 2y + 43y=2y+4

=>y = 4y=4

Now, you can use this y = 4y=4 to find the value of xx. Going back to our expression for xx:

=>x = 2y - 2x=2y2

=>x = 2(4) -2x=2(4)2

=>x = 8 - 2x=82

=>x = 6x=6

So the solution is x = 6x=6 and y = 4y=4.

Apr 13, 2018

y=4y=4
x=6x=6

Explanation:

x=2y-2x=2y2
3y=x+63y=x+6

The first one will be easiest to plug into the second, since it is already defined for xx. So, we will be substituting the xx in the second equation with the definition of xx provided in the first.

3y=(2y-2)+63y=(2y2)+6

The parenthesis do not prevent you from combining like terms. In this case, the -22 and 66.

3y=2y+43y=2y+4

3y-2y=43y2y=4

y=4y=4

Now plug the yy in to the top problem to solve for xx.

x=2(4)-2x=2(4)2

x=6x=6

Results: y=4y=4 and x=6x=6

Apr 13, 2018

x=6x=6
y=4y=4

Explanation:

x=2y-2x=2y2
3y=x+63y=x+6

To solve for each variable through substitution we have to make one equation have only one variable in it (either xx or yy). In order to do this, we have to put one variable in terms of the other. This time we got lucky because the first equation already puts xx in terms of yy. We just have to plug in 2y-22y2 for xx in the second equation.

3y=x+63y=x+6

3y=(2y-2)+63y=(2y2)+6

3y=2y+43y=2y+4

Subtract 2y2y from each side

y=4y=4

Now plug this value into one of the original equations to find xx.

x=2y-2x=2y2

x=2(4)-2x=2(4)2

x=8-2x=82

x=6x=6

OR

3y=x+63y=x+6

3(4)=x+63(4)=x+6

12=x+612=x+6

6=x6=x

Answer

x=6x=6
y=4y=4