Solving Using Substitution
| Site: | Gladwin |
| Course: | Michigan Algebra I Sept. 2012 |
| Book: | Solving Using Substitution |
| Printed by: | Guest user |
| Date: | Thursday, November 21, 2024, 1:35 PM |
Description
Substitution
When graphing is not a viable option, systems of equations can be solved symbolically. One method for a symbolic solution is substitution. When solving linear systems using substitution, one variable must be replaced with an equivalent expression that includes the other variable.
Example 1 Solve the following system by substitution.
Step 1. Isolate one variable in one equation.
Since y is isolated in the first equation, use that variable and that equation.
Step 2. Substitute the isolated variable's equivalent expression into the other equation.
Since y is equal to 2x + 4, substitute the "y" in 2x + 4y = 6, with the equivalent expression, 2x +4.
2x + 4(2x + 4) = 6
Step 3. Solve the remaining equation.
2x + 8x + 16 = 6
10x +16 = 6
10x = -10
x = -1
Step 4. Substitute the value of the variable found in step 3, into either equation to find the other variable.
y = 2(-1) + 4
y = -2 + 4
y = 2
Therefore the solution point is (-1, 2).
Example 2
Peter and Nadia like to race each other. Peter can run at a speed of 5 feet per second and Nadia can run at a speed of 6 feet per second. To be a good sport Nadia likes to give Peter a head start of 20 feet. How long does Nadia take to catch up with Peter? At what distance from the start does Nadia catch up with Peter?
Step 1. Write a linear equation to represent each person in the situation.
Nadia's equation d= 6t
Peter's equation d = 5t + 20
Step 2. Isolate one variable in one equation.
In this example both equations have d isolated.
Step 3. Substitute the isolated variable's equivalent expression into the other equation.
6t = 5t + 20
Step 4. Solve the remaining equation.
6t = 5t + 20
t = 20
Step 5. Substitute the value of the variable found in step 3, into either equation to find the other variable.
d = 6t
d = 6(20)
d = 120
Nadia and Peter meet 20 seconds after they start racing,
at a distance of 120 yards away.
Example 3
Solve the system
2x + 3y = 6
-4x + y = 2
Step 1. Isolate one variable in one equation.
Since the y in the second equation has a coefficient of one, it is the easiest to isolate.
-4x + 4x + y = 2 + 4x
y = 2 + 4x
Step 2. Substitute the isolated variable's equivalent expression into the other equation.
Since y is equal to 4x + 2, substitute the "y" in 2x + 3y = 6, with the equivalent expression, 4x +2.
2x + 3(4x + 2) = 6
Step 3. Solve the remaining equation.
2x + 12x + 6 = 6
14x + 6 = 6
14x = 0
x = 0
Step 4. Substitute the value of the variable found in step 3, into either equation to find the other variable.
y = 2 + 4(0)
y = 2
Therefore, the solution point is (0, 2).
Video Lesson
To see more examples on solving linear systems using substitution, select the following links:
Guided Practice
To solidify your understanding of solving systems using substitution, visit the following link to Holt, Rinehart and Winston Homework Help Online. It provides examples, video tutorials and interactive practice with answers available. The Practice and Problem Solving section has two parts. The first part offers practice with a complete video explanation for the type of problem with just a click of the video icon. The second part offers practice with the solution for each problem only a click of the light bulb away.
Practice
Systems by Substitution Worksheet
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Answer Key
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Sources
Andrew and Anne Gloag, This chapter covers solving systems of equations graphically, Created: Jan. 18, 2010, http://www.ck12.org/flexr/search/Linear%20systmes%20of%20Equations/
Holt, Rinehart, & Winston. "Multi Step Equations." http://my.hrw.com/math06_07/nsmedia/homework_help/msm3/
msm3_ch11_06_homeworkhelp.html (accessed August 3, 2010).