Transcript Solving Linear Systems

Solving Linear Systems
Substitution Method
Lisa Biesinger
Coronado High School
Henderson,Nevada
Linear Systems
A linear system consists of two or
more linear equations.
The solution(s) to a linear system is
the ordered pair(s) (x,y) that satisfy
both equations.
Example
Linear System:
x  2y  1
5 x  4 y  23
The solution will be
the values for x and
y that make both
equations true.
Solving the System
Step 1: Solve one of
the equations for either
x or y.
x  2y  1
5 x  4 y  23
For this system, the first
equation is easy to
solve for x because the
coefficient of x is equal
to 1.
Organizing Your Work
Set up 2 columns on your paper.
Place one equation in each column.
Write the equation we are using first in
column 1, and the other equation in
column 2.
Solving by Substitution
x  2y  1
Now we will solve
for x in column 1.
5 x  4 y  23
Solving by Substitution
x  2y  1
x  2 y  2 y 1 2 y
x  1 2 y
Subtract 2y from
both sides.
5 x  4 y  23
Solving the System
Step 2: Substitute
your answer into the
other equation in
column 2.
Substituting will
eliminate one of the
variables in the
equation.
Solving by Substitution
x  2y  1
x  2 y  2 y 1 2 y
x  1 2 y
5 x  4 y  23
5(1  2 y )  4 y  23
Always use
parenthesis when
substituting an
expression with two
terms.
Solving the System
Step 3: Simplify
the equation in
column 2 and solve.
Solving by Substitution
x  2y  1
x  2 y  2 y 1 2 y
x  1 2 y
5 x  4 y  23
5(1  2 y )  4 y  23
5  10 y  4 y  23
5  14 y  23
Use the distributive
property and combine
like terms.
Solving by Substitution
x  2y  1
x  2 y  2 y 1 2 y
x  1 2 y
5 x  4 y  23
5(1  2 y )  4 y  23
5  10 y  4 y  23
5  14 y  23
5  5  14 y  23  5
 14 y  28
y2
Solve for y.
Solving the System
Step 4: Substitute
your answer in
column 2 into the
equation in column
1 to find the value
of the other
variable.
Solving by Substitution
x  2y  1
x  2 y  2 y 1 2 y
x  1 2 y
x  2y  1
x  2 ( 2)  1
x4 1
x  4  4  1 4
x  -3
Substitute for y and solve
for x. Solve for x.
5 x  4 y  23
5(1  2 y )  4 y  23
5  10 y  4 y  23
5  14 y  23
5  5  14 y  23  5
 14 y  28
y2
The Solution
The solution to this linear system is
x  - 3 and y  2 .
The solution can also be written as an
ordered pair  3,2
Almost Finished
Checking Your Solution
Check your answer
by substituting for x
and y in both
equations.
Checking Your Answer
x  2y  1
 3  2(2)  1
3 4 1
11 
5 x  4 y  23
5(3)  4(2)  23
 15  8  23
 23  23

Additional Examples
Problem #1:
Problem#2
Problem #3
Answers:
#1
#2
#3