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Solving system of equations by substitution, also known as the substitution method, works by first solving each of the equations for one of the variables.

For example, consider the system x = 4y + 2, y = 11x + 2

The first equation is solved for x, and the second is solved for y by isolating each of the variables in the equation.

Once we have written our system in this way, we substitute the equation for one variable into the other equation in place of that variable. So we would plug the isolated expression for x into the variable x in the isolated equation for y.

We know x = 4y + 2, so that means y = 11(4y + 2) + 2, and now we can solve this equation for y.

Let’s look at a complete example:

What is the value of x and y for the system 2x – 5y = -7 and x + 3y = 13?

Let’s number our two equations:

1) 2x – 5y = -7

2) x + 3y = 13

We can rearrange equation 2 by moving the 3y to the right side, which gives us equation 3:

3) x = 13 – 3y

Now we can plug equation 3 into equation 1.

2x – 5y = -7

2(13 – 3y) – 5y = -7 This is plugging in 13 – 3y for “x” in equation 1

26 – 6y – 5y = -7

-11y = -33

y = 3

We’ve found y! We’ll call this equation 4:

4) y = 3

Now we can plug equation 4 into equation 1, 2 or 3 to solve for x. Let’s use equation 3:

x = 13 – 3y

x = 13 – 3(3)

x = 13 – 9

x = 4

And that’s it! Just as a sanity check, we can try plugging in equation 4 into equation 1:

2x – 5y = -7

2x – 5(3) = -7

2x – 15 = -7

2x = -7 + 15

2x = 8

x = 4

We get the same answer, which is a great sign that we’ve done things right.

Let’s do one more example:

What is the value of x and y for 3x + 2y = 11, and 7x – 4y = 10?

Let’s number our equations again:

1) 3x + 2y = 11

2) 7x – 4y = 10

We can beging with either equation, let’s rearrange equation 1:

3x + 2y = 11

3x = 11 – 2y

x = ^{11}/_{3} – ^{2}/_{3}y We’ll call this equation 3, x in terms of y

Now let’s plug equation 3 into equation 2:

7x – 4y = 10

7(^{11}/_{3} – ^{2}/_{3}y) – 4y = 10

^{77}/_{3} – ^{14}/_{3}y – 4y = 10

^{77}/_{3} – ^{14}/_{3}y – ^{12}/_{3}y = 10

^{77}/_{3} – ^{26}/_{3}y = 10

^{26}/_{3}y = ^{77}/_{3} – ^{30}/_{3}

^{26}/_{3}y = ^{47}/_{3}

y = ^{47}/_{26} Let’s call this equation 4

Pretty annoying fractions, but we’ve found y! Now let’s plug equation 4 into equation 1:

3x + 2y = 11

3x + 2(^{47}/_{26}) = 11

3x + ^{47}/_{13} = 11

3x = ^{143}/_{13} – ^{47}/_{13}

3x = ^{96}/_{13}

x = ^{33}/_{13}

And that’s it. If you want to make sure it’s right, we can plug in our answers into the original equations 1 and 2.

1) 3x + 2y = 11

2) 7x – 4y = 10

1) 3(^{33}/_{13}) + 2(^{47}/_{26}) = 11

2) 7(^{33}/_{13}) – 4(^{47}/_{26}) = 10

If you find that this doesn’t hold true, it means we made an error somewhere.

The best way to learn is by practicing, so try out these examples on your own:

What is the value of x and y for 2x + 4y = 22 and 7x – 3y = 26?

Answer: x = 5, y = 3

What is the value of x and y for ^{1}/_{2}x – 6y = -1 and 8x + 3y = 17?

Answer: x = ^{1}/_{2}, y = ^{1}/_{3}

If you can solve these, you’re all set for solving system of equations by substitution!

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