Topic: Solving Systems of Linear Equations Algebraically
Solving Linear Systems Algebraically
 Substitution Method
 Isolate a variable in one of the equations (doesn’t matter which equation
or which variable).
 Substitute the result into the other equation & solve for the variable in
question.
 Use that solution to substitute and solve for the other variable.
 Elimination (Linear Combination) Method
 Combine the two equations by adding or subtracting them, creating a
single equation in which one variable has been eliminated, making it easy
to solve for the other variable.
 Like terms & “=“ signs must be aligned.
 Each equation should have the same coefficient in front of one of the
variables.
 Use that solution to substitute and solve for the other variable.
Solving Linear Systems: Substitution
2x  y  8
y  x7
{
The variable y is already isolated in the 2nd equation. Substitute
the right side of that equation for y in the 1st equation; always
protect your substitution with ( ).
2x  y  8
2 x  ( x  7)  8
Now solve this equation for x.
3x  7  8
3 x  15
x5
Substitute this value into one of the
original equations and solve for y; it
doesn’t matter which equation, but to
keep things simple we’ll use the 2nd
one.
y  x7
y  57
y  2
The solution to the system is the
point (5, -2).
Solving Linear Systems: Substitution
y x3 0
x  y3
{
The variable x is already isolated in the 2nd equation. Substitute
the right side of that equation for x in the 1st equation;
remember to protect your substitution with ( ).
y x3 0
y  ( y  3)  3  0
Now solve this equation for y.
y  y 33  0
00
Our variable disappeared and left us with a true statement, meaning this equation is an
identity and is always true. Therefore, our system is dependent and has infinite
solutions.
Solving Linear Systems: Elimination
3 x  y  26
2 x  y  19
{
Each y term has the same coefficient. Since one is positive and
one is negative, we can add the two equations together to
eliminate y.
3 x  y  26
 (2 x  y  19)
5x
 45
Now solve this equation for x.
5 x  45
x  9
Substitute this value into one of the
original equations and solve for y. Since y
is positive in the 1st equation, that’s
probably the easier one to simplify.
3(9)  y  26
 27  y  26
y  1 The solution to the system is the
point (-9, 1).
Solving Linear Systems: Elimination
x  3 y  17
2 x  4 y  20
{
Neither equation has the same coefficient in front of one of the
variables. However, we can simplify the 2nd equation by a
factor of 2, which would give both x terms the same coefficient.
(2 x  4 y  20)  2
x  2 y  10
Solving Linear Systems: Elimination
x  3 y  17
x  2 y  10
{
Now we can combine the equations to eliminate x. Since both
coefficients are the same sign, we must subtract (WATCH
YOUR SIGNS!)
x  3 y  17
 ( x  2 y  10)
y7
That was easy! Now we substitute this into one of the original
equations and solve for x.
x  3(7)  17
x  21  17
The solution to the system is the point (-4, 7).
x  4
How do I know what method to use?
 Graphing
 Both equations are already in slope-intercept form, or can easily
be rewritten in slope-intercept.
 Substitution
 One of the equations already has a variable isolated, or a
variable can be easily isolated.
 Elimination
 One of the variables has the same coefficient in both equations,
or one equation can be easily divided or multiplied to get the
same coefficient.
 REMEMBER: Each method is equally valid. The goal is to
select the most efficient method depending on the
system given.
JOURNAL ENTRY
 TITLE: Checking My Understanding: Solving Systems of
Linear Equations Algebraically
 Review your notes from this presentation & create and
complete the following subheadings in your journal:
 “Things I already knew:” Identify any information with which
you were already familiar.
 “New things I learned:” Identify any new information that you
now understand.
 “Questions I still have:” What do you still want to know or do
not fully understand?
Homework
 Quest: Solving Systems of Equations Algebraically
 Due 1/23 (A-day) or 1/24 (B-day)