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*y* = 3 into equation (3),

When you have worked them out, check that you got*x* = 7 and
*y* = 4.

There is a more advanced type of substitution, which shall be left to a later algebra lesson.

More Lessons for Algebra

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This Algebra Lesson introduces the technique of substitution - another key technique in algebra.

Substitution works this way: if an unknown quantity (or an expression) can be expressed in a different way, then you can substitute in its place the alternative expression. This gives you a new equation, which may lead to the solution.

Let's start with a simple example to illustrate substitution. Suppose you are given:

3x+y= 30

and

3x= 24

Since we know that 3*x* = 24 we can substitute
24 in the place of
3*x* into the first equation to give us:

24 +y= 30

Then we can solve for *y* as follows:

24 +y− 24 = 30 − 24

y= 30 − 24

∴y= 6

And we can readily solve for *x* from the second equation:
3*x* = 24 to get *x* = 8 which should
be straightforward (if you need help with that, you may want to review our lessons on
Basic Algebra or Transposition).

Why didn't we just solve for *x* first and plug 8
into the first equation? Of course we
could've done that, but we just wanted to illustrate the substitution technique using a simple
example first.

Now, for a more typical example to develop our understanding of substitution. If you have two unknown
quantities, *x* and *y*, and two
equations linking them. (Two equations are needed to get a unique solution for two unknowns. If there
was just one equation, an infinite number of pairs of *x* and
*y* would exist that satisfy it):

x+ 3y= 13 (1)

2x–y= 5 (2)

We can subtract 3*y* from both sides of equation (1):

x= 13 – 3y(3)

Now we substitute this for *x* into equation (2)

2(13 – 3y) –y= 5

That gives us:

26 – 6y–y= 5

Now, subtract 26 from both sides.

– 6y–y= – 21

– 7y= – 21

Multiply both sides by (–1).

7y= 21

We now solve for *y*

Then substitutey= 3

x= 13 – 3y

x= 13 – 3 × 3

x= 13 – 9

x= 4

To check our answers, put *x* = 4 and
*y* = 3 into equations (1) and (2) to see that
these solutions indeed satisfy the equations.

With this algebra lesson, you've completed the introduction to the technique of substitution in algebra.

Now, try to solve the following yourself, using the steps shown above.

x−y= 3 (1)

2x+ 3y= 26 (2)

When you have worked them out, check that you got

There is a more advanced type of substitution, which shall be left to a later algebra lesson.

If you would like more practice on this technique, try our Solving System of Equations by Substitution worksheet.

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