Lesson 16 Solve Systems of Equations Algebraically
Lesson 16Solve Systems of Equations Algebraically
You know that solutions to systems of linear equations can be shown in graphs. Now you will learn about other ways to find the solutions. Take a look at this problem.
Sienna wrote these equations to help solve a number riddle.
y 5 x 2 20
x 1 y 5
84
What values for x and y solve both equations?
Use math you already know to solve the problem.
a. What does y 5 x 2 20 tell you about the relationship between x and y?
b. What does x 1 y 5 84 tell you about the relationship between x and y?
c. You can guess and check to solve the problem. Try 44 for x and 40 for y. Do these numbers solve both equations?
d. Now try 50 for x. If x 5 50, what is y when y 5 x 2 20? Does that work with the other equation?
e.
Try x 5 52. What do you find?
f. Explain how you could find values for x and y that solve both equations.
Lesson 16 Solve Systems of Equations Algebraically
In Lesson 15, you learned that without actually solving, you can tell if a system of equations will have exactly one solution, no solution, or infinitely many solutions. Here are some examples.
x 1 y 5 6 2x 1 2y
5 12
The second equation is double the first one, so they are equivalent equations with the same graph and solution set. This system has infinitely many solutions.
5x 1 y 5 3 y 5 4 2 5x
If you write both equations in slope-intercept form, you find that y 5 25x 1 4 and y 5 25x 1 3. The lines have the same slope and different y-intercepts so they are parallel. This system has no solution.
If a system of equations has exactly one solution, like the problem on the previous
page, there are different ways you can find the solution.
You could guess and check, but that is usually not an efficient way to solve a system of equations. You could graph each equation, but sometimes you can’t read an exact answer from the graph. Here is one way to solve the problem algebraically.
y 5 x 2 20 x 1 y 5 84
Substitute x 2 20 for y in the second equation and solve for x. x 1 (x 2 20) 5 84 2x 2 20 5 842x 5 104 x 5 52, so y 5 32
You will learn more about
algebraic methods later in the lesson.
Reflect1 How does knowing x 5 52 help you find the value of y?
Lesson 16 Solve Systems of Equations Algebraically
Using Substitution to Solve Systems of Equations
Read the problem below. Then explore how to use substitution to solve systems of equations.
Solve this system of equations.
y
5 x 1 2
y 1 1 5 24x
Graph It You can graph the equations and estimate the solution.
Find the point of intersection. It looks like
the solution is close to (2 1 ·· 2 , 1 1 ·· 2 ) .
Model It You can use substitution to solve for the first variable.Notice that one of the equations tells you that y 5 x 1 2. This allows you to use substitution to solve the system of equations.
Connect It Now you will solve for the second variable and analyze the solution.2 What is the value of x? How can you find the value of y if you know the value of x?
3
Substitute the value of x in the equation y 5 x 1 2 to find the value of y.
4 Now substitute the value of x in the equation y 5 24x 2 1 to find the value of y.
5 What is the ordered pair that solves both equations? Where is this ordered pair located
on the graph?
6 Look back at Model It. How does substituting x 1 2 for y in the second equation give you
an equation that you can solve?
7 How does substitution help you to solve systems of equations?
Try It
Use what you just learned to solve these systems of equations. Show your work on a separate sheet of paper.
Lesson 16 Solve Systems of Equations Algebraically
Using Elimination to Solve Systems of Equations
Read the problem below. Then explore how to solve systems of equations
using elimination.
Solve this system of equations.
2x 2 2y 5 4
3y 5 20.5x 1 2
Model It You can use elimination to solve for one variable.First, write both equations so that like terms are in the same position. Then try to eliminate one of the variables, so you are left with one variable. To do this, look for a way to get opposite coefficients for one variable in the two equations.
22y 5 x 1 4
3y 5 20.5x 1 2
2(3y 5 20.5x 1 2)6y 5 2x 1 4
22y 5 x 1
46y 5 2x 1 44y 5 8
y 5 2
2x 2 2(2) 5 42x 2 4 5 4
2x 5 8x 5 28
The solution is x 5 28, y 5 2.
Check:
3(2) 5? 20.5(28) 1 2
6 5 4 1 2
• Multiply the second equation by 2. Now you have opposite terms: x in the first equation and 2x in the second equation.
• Add the like terms in the two equations. The result is an equation in just one variable.
• Divide each side by 4 to solve the equation for y.
• Substitute the value of y into one
of the original equations and solve for x.
• Substitute your solution in the other original equation.
Lesson 16 Solve Systems of Equations Algebraically
Solving Systems of Equations Algebraically
Solve the problems.
1 Which statement about this system of equations is true?
y 5 1 ·· 2 (8x 1 4)
y 2 4x 5 2
A
It has no solution. C (0, 2) is the only solution.
B It has exactly one solution. D It has infinitely many solutions.
2 For each pair of linear equations, shade in the box under the appropriate column to indicate whether the pair of equations has no solution, exactly one solution, or infinitely many solutions.
Equations No Solution Exactly One Solution Infinitely Many Solutions
3x 1 4y 5 5 3x 2 4y 5 1
y 5 2x 1 6 y 5 2x 2 9
x 2 3y 5 7 2x 2 6y 5 14
y 5 x
6x 2 y 5 4
3 The graphs of lines a and b are shown on the coordinate plane. Match each line with its equation. Write the equation on the line provided for each line. Then solve the system of equations using any method.
Go back and see what you can check off on the Self Check on page 99.
4 You know that the only solution of a system of two linear equations is (22, 1). Graph two lines that could be in this system of equations. Label the lines a and b. What has to be true about the graphs of both lines? What has to be different about the
lines?
Show your work.
5 What is the equation of each line you drew in Problem 4? Explain how to check algebraically that your system of equations has a solution of (22, 1).
