*Substitution*", so to do substitution, you would first solve one of your equations for one of the variables (let's say, the second equation, and that we will solve for x).

x+3y=4

2x+3y=2

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Quang H. | Math and Science TutorMath and Science Tutor

You would take one equation and subtract it from the other like this:

2x + 3y = 2

- x + 3y = 4

______________

x + 0 = -2

So we found that x = -2. We can then plug in -2 into x for either one of the equations and solve for y:

(-2) + 3y = 4

3y = 6

y = 2

Although, you have, as your topic, "Solving Systems By *Substitution*", so to do substitution, you would first solve one of your equations for one of the variables (let's say, the second equation, and that we will solve for x).

Taking the second equation and solving for x, we get:

x = 4 - 3y

You can then plug in what you have just found as x into the other equation like this:

2x + 3y = 2

2(4 - 3y) + 3y = 2

Solving the equation for y, we get:

2(4 - 3y) + 3y = 2

8 - 6y + 3y = 2

8 - 3y = 2

-3y = -6

y = 2

And then you can plug in this value for y into any one of the first equations to solve for x.

x+3y=4

2x+3y=2

by substitution...

x+3y=4, x=4-3y

2x+3y=2

2(4-3y)+3y=2

8-6y+3y=2

8-3y=2

-3y=2-8

-3y=-6

y=2

x+3y=4

x+3(2)=4

x+6=4

x=4-6

x=-2

(-2,2) is the solution

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