1. Learning
Algebra
Solving Linear Systems of
Equations
By:
Mr. Patrick Paez

2. The first thing we need to
do is learn about what
systems of linear
equations are.
Systems of linear
equations are two or more
linear equations of the
same variables.
A point that lies on both
lines would be the solution
to the system.

3. Let’s look at an example of a system of linear equations.
x + 2y = 7
3x – 2y = 5
Since we are just dealing with an x – y coordinate system, or a
Cartesian coordinate system, linear equations will either
intersect or not intersect, making them parallel.
If they do intersect, the point of intersection is the solution to
the system.

4. There are three different
ways to find solutions for
these systems of linear
equations. They are:
• Graphing
• Substitution
• Elimination

5. Let’s start with solving by graphing. As
stated earlier, the solution of two linear
equations is the point of intersection.
This point can be easily found by
following three simple steps.
1) Put both equations in y-intercept form.
2) Graph each equation and estimate the point
of intersection.
3) Check the coordinates algebraically by
substituting into the original equations.
GRAPHING

6. Here is an example of solving a system of linear
equations by graphing.
You have the following two equations. Use a graph
to solve the system of equations.
Eq 1: x + 2y = 7
x + 2y = 7
-x -x
2y = -x + 7
/2 /2 /2
y = -(1/2)x + 7/2
Eq 2: 3x – 2y = 5
3x - 2y = 5
-3x -3x
-2y = -3x + 5
/-2 /-2 /-2
y = (3/2)x - 5/2

7. Now let’s see how this looks on a graph
As we can see on the
graph, the
intersection is (3,2)
y = -(1/2)x + 7/2
y = (3/2)x - 5/2

8. The point, or coordinate, (3,2) would be the solution to that
particular system of equations. Remember to substitute the
point (3,2) into the original equations to check your answer.
(3) + 2(2) = 7
3 + 4 = 7
7 = 7
3(3) – 2(2) = 5
9 – 4 = 5
5 = 5
Now, using the graphing method, try to find the solution
for the following systems of equations.
Problem 1
-x + y = -7
x + 4y = -8
Problem 2
y = -x + 4
y = 2x - 8

9. Problem 1
The two lines
intersect at
point (4,-3)
Don’t forget to
check your
answer:
y = x - 7
y = -(1/4)x - 2
-x + y = -7
-(4) + (-3) = -7
-7 = -7
x + 4y = -8
(4) + 4(-3) = -8
-8 = -8

10. Problem 2
The two lines
intersect at
point (4,0)
Don’t forget to
check your
answer:
y = -x + 4
0 = -(4) + 4
0 = 0
y = 2x - 8
0 = 2(4) - 8
0 = 0

11. Substitution
Graphing is a good way to solve these
systems. However, there will be some
problems where graphing will get a little
difficult. Good thing there is another
strategy you can use. Here are the
steps for solving by substitution.
1) Solve one of the equations for one variable.
2) Substitute the expression from step 1 into
the other equation and solve for the other
variable.
3) Substitute the value from step 2 into the
revised equation from step 1 and solve.

12. Example 1:
y = 3x + 2
x + 2y = 11
Step 1: Solve one equation for a variable.
Equation 1 is already solved for y.
Step 2: Substitute 3x + 2 for y in equation 2
and solve for x.
x + 2y = 11
x + 2(3x + 2) = 11
x + 6x + 4 = 11
7x = 7
x = 1

13. Example 1 (continued):
Step 3: Substitute 1 for x in the original
equation 1 to find the value of y.
y = 3x + 2
y = 3(1) + 2
y = 5
Finish your work by checking your answers
in equation 2.
x + 2y = 11
1 + 2(5) = 11
1 + 10 = 11
11 = 11

14. Elimination
Again, some equations may be a little
harder to solve when using graphing or
substitution. Solving by elimination can
be one of the easier methods to use
given the right situation. These are the
three steps.
1) Add or subtract the equations to eliminate
one variable.
2) Solve the resulting equation for the other
variable.
3) Substitute in either original equation to find
the value of the eliminated variable.

15. 2x + 3y = 11
-2x + 5y = 13
Step 1: Add the equations to eliminate one
variable.
2x + 3y = 11
-2x + 5y = 13
8y = 24
Step 2: Solve for y
8y = 24
y = 3
Example 1:

16. Example 1 (continued):
Step 3: Substitute 3 for y in either equation
and solve for x.
2x + 3y = 11
2x + 3(3) = 11
2x + 9 = 11
2x = 2
x = 1
Finish your work by checking your answers
in equation 2.
-2x + 5y = 13
-2(1) + 5(3) = 13
-2 + 15 = 13

17. Elimination: Part 2
Not all equations may be set up as nicely
as the previous example. In some
equations, you may have to multiply an
equation by a constant so you can add
or subtract the equations to eliminate
one variable.

18. Example 2:
6x + 5y = 19
2x + 3y = 5
Step 1: Multiply the bottom equation by -3 to
make eliminating x possible.
6x + 5y = 19
-3(2x + 3y = 5)
6x + 5y = 19
-6x – 9y = -15
Step 2: Add the two equations together to
eliminate x.
-4y = 4

19. Example 2 (continued):
Step 3: Solve for y
-4y = 4
y = -1
Step 4: Substitute -1 for y in either equation
and solve for x.
6x + 5y = 19
6x + 5(-1) = 19
6x = 24
x = 4

20. Finish your work by checking your
answers in equation 2.
2x + 3y = 5
2(4) + 3(-1) = 5
8 - 3 = 5
5 = 5
Thank you for your time…and don’t
forget to study!!!