In this presentation we’re going to learn about the substitution method.

1. How to solve linear
equations by substitution
Welcome to Mr. JD’s EL Math class.

2. In this presentation we’re going to learn about the
substitution method.

3. If you look up Algebra 1, Chapter 6, System of equations
and inequalities, page 404, you’ll probably see two
equations that looks something like this.
Y=2X
Y=X+5

4. The idea here is to solve one of the equations for one of the
variables, and plug this into the other equation. It does not matter
which equation or which variable you pick. There is no right or
wrong choice; the answer will be the same, regardless. But —
some choices may be better than others

5. For instance, in this case, can you see that it would probably be
simplest to solve the first equation for "y =", since there is
already a y value there? I could solve the second equation for
either variable, then I'll solve the second equation for y:

6. Step 1
Y= 2 X
Y=X+5
Both equations are solved for
Y
Step 2
Y= 2 X
2X=X+5
Substitute 2X for Y in the
second equation
Step 3
2X = X + 5
-X -X
Solve for X, by subtracting X
from both sides
X = 5 Rewrite
X = 5
Step 4
Y = 2X “X” equals 5
Y = 2 (5) In the following step you will
substitute X and enter its value.
Y = 10 “Y” equals 10
Step 5
(5,10) Now write the solution as an
ordered pair.

7. Wow! That’s kind of overwhelming when you see the number of
steps all at once! But don’t worry… we’ll take them one “step” at
a time and you’ll see this method is not that complicated after all.
In fact, once you completed the steps, and if you want to double
check your work, you can do the following:

8. Step 1.
Double check, by entering both equations again. Keep the coordinates “X” and “Y” (5,10).
Y=2X Y= X + 5
Step 2
Substitute by entering the coordinates instead of the variables “X” and “Y”
10 = 2 (5) 10= 5 + 5
10 = 10 10 = 10
Ten is equal to the product of 2 times 5. So, this is
true
Ten is equal to the sum of 5 plus 5. So, this is true

9. That means we’ve pretty much got things covered, right?
Ah… not so fast! We need to practice with new equations, so
let’s do it!

10. Solve each system by substitution
2x + y = 5
y = x - 4
Step 1
y = x - 4 The second equation is solved for y.
Step 2
Step 2 2x + y = 5
2x + (x - 4) = 5
Substitute x - 4 for y in the first equation.
Step 3
3x - 4 = 5
+ 4 +4
Simplify. Then solve for x.
Add 4 to both sides.
3x = 9
3𝑋
3
=
9
3
Divide both sides by 3.
X = 3
Now that we have the value of “X”, lets continue
working to find the value of Y
Y = X - 4 Write one of the original equations.
Y = 3 - 4
Substitute 3 for x.
Y = - 1
Write the solution as an ordered pair. As follows: (X,Y) or (3,-1), now
lets double check your work, by entering any of the original equations.
2X + Y = 5 Y= X - 4
2(3) + (-1) = 5 -1= 3 - 4
6 + (-1) = 5 -1 = -1
5 = 5 -1 = -1
Lets find the value of “X”

11. How about one more problem and then we move to how to
solve linear equations by elimination…..so let’s go!

12. Solve each system by substitution
2x + y = 5
y = x - 4
Step 1
y = x - 4 The second equation is solved for y.
Step 2
Step 2 2x + y = 5
2x + (x - 4) = 5
Substitute x - 4 for y in the first equation.
Step 3
3x - 4 = 5
+ 4 +4
Simplify. Then solve for x.
Add 4 to both sides.
3x = 9
3X
3
=
9
3
Divide both sides by 3.
X = 3
Now that we have the value of “X”, lets continue working to
find the value of Y
Y = X - 4 Write one of the original
equations.
Y = 3 - 4
Substitute 3 for x.
Y = - 1
Write the solution as an ordered pair. As follows: (X,Y) or (3,-
1), now lets double check your work, by entering any of the
original equations.
2X + Y = 5 Y= X - 4
2(3) + (-1) = 5 -1= 3 - 4
6 + (-1) = 5 -1 = -1
5 = 5 -1 = -1
Lets find the value of “X”

13. Now let’s try by yourselves?

14. Solving Systems of Equations by Substitution
Solve each system by substitution. (To make it easy for you, we have attached a link at
schoology so you could print it)
y= 6x – 11
−2x − 3y = −7
2x− 3y = −1
y= x – 1
y= −3x + 5
5x − 4y = −3
−3x − 3y = 3
y = −5x – 17
y = −2
4x − 3y = 18
y = 5x – 7
−3x − 2y = −12
−4x + y = 6
−5x − y = 21
−7x − 2y = −13
x − 2y = 11
−5x + y = −2
−3x + 6y = −12
−5x + y = −3
3x − 8y = 24

15. Alright, so now you know how to solve linear equations by substitution, we
will proceed to the elimination method. Obviously there’s a good chance
that you’ll see them explained in slightly different forms, or different
orders, or even using different terminology in other math classes or books,
but the basic ideas will be the same. Please, follow your teacher classes
very closely!
And what’s the best way to understand how to solve linear equations by
substitution or elimination?
Yep, ya gotta practice! So be sure to do some problems with msubstitution
method on your own.
As always, thanks for watching my presentation and I’ll see ya next time.
Mr. Delgado