Vocabulary
Substitution
Solving by Substitution
Systems of linear equations do not have to be graphed in order to be solved. Systems of linear equations can be solved algebraically. One method of solving a system algebraically is to use a process called substitution.
Substitution has three steps:
Substitution has three steps:
- When necessary, solve at least one equation for one variable.
- Substitute the equation you just found in for the same variable in the other equation. Solve this equation.
- Substitute the answer you just found into either original equation. Write the answer as an ordered pair
Examples
Example 1: Solve the system of linear equations by using substitution.
y = 2x + 1
3x + y = -9
Step 1: Notice that y = 2x + 1 is already solved in terms of a variable, y. This step is done.
Step 2: Since y = 2x + 1, we can substitute in 2x + 1 for y in the second equation.
- 3x + y = -9 Original equation.
- 3x + (2x + 1) = -9 Substitute 2x + 1 in for y.
- 5x + 1 = -9 Combine like terms.
- 5x = -10 Subtract 1 from both sides.
- x = -2 Divide both sides by 5.
- y = 2x + 1 Original equation.
- y = 2(-2) + 1 Substitute -2 in for x.
- y = -4 + 1 Multiply.
- y = -3 Add.
As a review from the lesson before, since there is a solution to this system, it is consistent and independent.
Example 2: Solve the system of linear equations by using substitution.
x + 2y = 6
3x - 4y = 28
Step 1: This problem differs from Example 1 because there is not an equation that has an equation solved for a variable. Therefore, we need to pick an equation to rewrite. The top equation would be the better option, because the variable x is almost solved for and we can finish the job. However, any equation can be rewritten in terms of either variable--take your pick. But, for this example, we will rewrite x + 2y = 6.
- x + 2y = 6 Original equation.
- x = -2y + 6 Subtract 2y from each side.
- 3x - 4y = 28 Original equation.
- 3(-2y + 6) - 4y = 28 Substitute -2y + 6 in for x.
- -6y + 18 - 4y = 28 Distribute.
- -10y + 18 = 28 Combine like terms.
- -10y = 10 Subtract 18 from each side.
- y = -1 Divide both sides by -10.
- x = -2y + 6 The top equation, rewritten.
- x = -2(-1) + 6 Plug -1 in for y.
- x = 2 + 6 Multiply.
- x = 8 Add.
As a review from the lesson before, since there is a solution to this system, it is consistent and independent.
Example 3: Solve the system of linear equations by using substitution.
2x + 2y = 8
x + y = -2
Step 1: An equation will have to be rewritten. x + y = -2 will be the easiest to manipulate. It doesn't matter which variable is solved for, it will be the same amount of work.
- x + y = -2 Original equation.
- y = -x - 2 Subtract x from each side.
Step 2: Since y = -x - 2, substitute -x - 2 in for y in the other equation.
- 2x + 2y = 8 Original equation.
- 2x + 2(-x - 2) = 8 Substitute -x - 2 in for y.
- 2x -2x - 4 = 8 Distribute.
- -4 = 8 Combine like terms; 2x and -2x cancel each other out.
We ended up with -4 = 8. This statement is false; therefore, no value of x or y can make this work. This means there is no solution.
As a review from the lesson before, since there is no solution to the system, it is an inconsistent system.
Example 4: Solve the system of linear equations by using substitution.
y = 2x - 4
3y - 6x = -12
Step 1: The first equation, y = 2x - 4, is already solved for in terms of y.
Step 2: Since y = 2x - 4, we can substitute 2x - 4 in for y in the other equation.
- 3y - 6x = -12 Original equation.
- 3(2x - 4) - 6x = -12 Substitute 2x - 4 in for y.
- 6x - 12 - 6x = -12 Distribute.
- -12 = -12 Combine like terms; 6x and -6x cancel each other out.
We ended up with -12 = -12. This is a true statement; therefore, any value of x or y will make this system true. This means that there are infinitely many solutions.
As a review from the lesson before, since there are infinitely many solutions, this system is consistent and dependent.
Application
A store sold a total of car stereo systems and speakers in one week. The stereo systems sold for $104.95 and speakers sold for $18.95. The sales from these two items totaled $6926.75. How many of each item were sold?
Define two new variables: c = number of car stereo systems sold, s = number of speakers sold
Profit equation: 104.95c + 18.95s = 6926.75
Total amount sold equation: c + s = 125
Step 1: Manipulate c + s = 125 to get a variable by itself.
Step 2: Substitute 125 - s in for c in the other equation.
Step 3: Take s = 72 and plug 72 in for s into either equation. The easiest one would be to plug it into c = 125 - s.
Therefore, 53 car stereo systems and 72 speakers were sold.
Define two new variables: c = number of car stereo systems sold, s = number of speakers sold
Profit equation: 104.95c + 18.95s = 6926.75
Total amount sold equation: c + s = 125
Step 1: Manipulate c + s = 125 to get a variable by itself.
- c + s = 125 Original equation
- c = 125 - s Subtract s from each side.
Step 2: Substitute 125 - s in for c in the other equation.
- 104.95c + 18.95s = 6926.75 Original equation.
- 104.95(125 - s) + 18.95s = 6926.75 Substitute 125 - s in for c.
- 13,118.75 - 104.95s + 18.95s = 6926.75 Distribute.
- 13,118.75 - 86s = 6926.75 Combine like terms.
- -86s = -6192 Subtract 13,118.75 from both sides.
- s = 72 Divide both sides by -86.
Step 3: Take s = 72 and plug 72 in for s into either equation. The easiest one would be to plug it into c = 125 - s.
- c = 125 - s Original equation.
- c = 125 - (72) Plug 72 in for s.
- c = 53 Subtract.
Therefore, 53 car stereo systems and 72 speakers were sold.
Summary
Substitution is the process of solving a system where an equation is rewritten in terms of one variable and that expression is substituted in for that variable into the other equation. This technique is very useful in all fields of math, where the value of a variable can only be solved if another equation is manipulated. This idea of substitution can be applied to non-linear equations as well.
Missouri Learning Standards
A.CED.3 Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context.
A.REI.6 Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables.
A.REI.6 Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables.
Mathematical Practices
(2) Reason abstractly and quantitatively.