# 03.2 Solving Linear Systems Algebraically - Winterrowd-math ```3.2 Solving Linear Systems
Algebraically
• What are the steps to solve a system by substitution?
• What clue will you see to know if substitution is a
good choice?
• What are the steps to solve a system by linear
combination?
• How many solutions are possible for a linear
system?
USING ALGEBRAIC METHODS TO SOLVE SYSTEMS
In this lesson you will study two algebraic methods for solving
linear systems. The first method is called substitution.
THE SUBSTITUTION METHOD
1
Solve one of the equations for one of its variables.
2
Substitute expression from Step 1 into other equation
and solve for other variable.
3
Substitute value from Step 2 into revised equation
from Step 1. Solve.
The Substitution Method
Solve the linear system using
the substitution method.
3 x + 4y  – 4
x + 2y  2
Equation 1
Equation 2
SOLUTION
Solve Equation 2 for x.
x + 2y  2
Write Equation 2.
x  – 2y + 2
Revised Equation 2.
Substitute the expression for x into Equation 1 and solve for y.
3x + 4y  – 4
3(– 2y + 2) + 4y  – 4
y5
Write Equation 1.
Substitute – 2y + 2 for x.
Simplify.
The Substitution Method
Solve the linear system using
the substitution method.
3 x + 4y  – 4
x + 2y  2
Equation 1
Equation 2
Substitute the expression for x into Equation 1 and solve for y.
3x + 4y  – 4
3(– 2y + 2) + 4y  – 4
y5
Write Equation 1.
Substitute – 2y + 2 for x.
Simplify.
Substitute the value of y into revised Equation 2 and solve for x.
x  – 2y + 2
Write revised Equation 2.
x  – 2(5) + 2
Substitute 5 for y.
x  –8
Simplify.
The solution is (– 8, 5).
The Substitution Method
Solve the linear system using
the substitution method.
CHECK
3 x + 4y  – 4
x + 2y  2
Equation 1
Equation 2
Check the solution by substituting back into the original equation.
3x + 4y  – 4
?
3 (– 8) + 4 (5)  – 4
–4  –4
Write original equations.
Substitute x and y.
Solution checks.
x + 2y  2
?
– 8 + 2 (5)  2
22
USING ALGEBRAIC METHODS TO SOLVE SYSTEMS
In the first step of the previous
example, you could have solved for either x or y in either
Equation 1 or Equation 2. It was easiest to solve for x in
Equation 2 because the x-coefficient was 1. In general you
should solve for a variable whose coefficient is 1 or –1.
CHOOSING A METHOD
If neither variable has a coefficient of 1 or –1, you can still use
substitution. In such cases, however, the linear combination
method may be better. The goal of this method is to add the
equations to obtain an equation in one variable.
USING ALGEBRAIC METHODS TO SOLVE SYSTEMS
THE LINEAR COMBINATION METHOD
1
Multiply one or both equations by a constant to
obtain coefficients that differ only in sign for one
of the variables.
2
Add revised equations from Step 1. Combine like
terms to eliminate one of the variables. Solve for
remaining variable.
3
Substitute value obtained in Step 2 into either
original equation and solve for other variable.
The Linear Combination Method: Multiplying One Equation
Solve the linear system using the
linear combination method.
2 x – 4y  13
4 x – 5y  8
Equation 1
Equation 2
SOLUTION
Multiply the first equation by – 2 so that x-coefficients differ only in
sign.
2 x – 4y  13
•
–2
4 x – 5y  8
and solve for y.
– 4x + 8y  – 26
4 x – 5y  8
3y  –18
y  –6
The Linear Combination Method: Multiplying One Equation
2 x – 4y  13
4 x – 5y  8
Solve the linear system using the
linear combination method.
Add the revised equations and solve for y.
Equation 1
Equation 2
y  –6
Substitute the value of y into one of the original equations.
CHECK
Write Equation 1.
2
x
–
4y

13
You can check the solution algebraically using
the method shown in the previous example.
Substitute – 6 for y.
2 x – 4(– 6)  13
2 x + 24  13
x–
The solution is
(
–
Simplify.
11
2
11
, –6
2
Solve for x.
).
The Linear Combination Method: Multiplying Both Equations
Solve the linear system using the
linear combination method.
7 x – 12 y  – 22
– 5 x + 8 y  14
Equation 1
Equation 2
SOLUTION
Multiply the first equation by 2 and the second equation by 3
so that the coefficients of y differ only in sign.
7 x – 12 y  – 22 •
2
14 x – 24y  – 44
– 5 x + 8 y  14
3
– 15 x + 24y  42
•
solve for x.
–x  –2
x 2
The Linear Combination Method: Multiplying Both Equations
Solve the linear system using the
linear combination method.
7 x – 12 y  – 22
– 5 x + 8 y  14
Add the revised equations and solve for x.
Equation 1
Equation 2
x2
Substitute the value of x into one of the original equations.
Solve for y.
– 5 x + 8 y  14
– 5 (2) + 8y  14
y=3
Write Equation 2.
Substitute 2 for x.
Solve for y.
The solution is (2, 3). Check the solution algebraically or graphically.
Linear Systems with Many or No Solutions
Solve the linear system
x – 2y  3
2x – 4y  7
SOLUTION
Since the coefficient of x in the first equation is 1, use
substitution.
x – 2y  3
x  2y + 3
Solve the first equation for x.
Linear Systems with Many or No Solutions
Solve the linear system
x  2y + 3
x – 2y  3
2x – 4y  7
Solve the first equation for x.
Substitute the expression for x into the second equation.
2x – 4y  7
Write second equation.
2(2 y + 3) – 4 y  7
Substitute 2 y + 3 for x.
67
Simplify.
Because the statement 6 = 7 is never true, there is no solution.
Linear Systems with Many or No Solutions
6 x – 10 y  12
– 15 x + 25y  – 30
Solve the linear system
SOLUTION
Since no coefficient is 1 or –1, use the linear combination
method.
6 x – 10 y  12
•
5
30 x – 50 y  60
– 15 x + 25 y  – 30 •
2
– 30 x + 50 y  –60
00
Because the equation 0 = 0 is always true, there are
infinitely many solutions.