Download Solving Equations by Graphing 5.5 and more Study notes Algebra in PDF only on Docsity!
Section 5.5 Solving Equations by Graphing 261
5.5 Solving Equations by Graphing
Solving an Equation by Graphing
Work with a partner. Solve 2 x − 1 = − —^1 2 x^ +^ 4 by graphing. a. Use the left side to write a linear equation. Then use the right side to write another linear equation. b. Graph the two linear equations from part (a). Find the x -value of the point of intersection. Check that the x -value is the solution of
2 x − 1 = − —^1 2 x^ +^ 4.
c. Explain why this “graphical method” works.
Essential QuestionEssential Question How can you use a system of linear
equations to solve an equation with variables on both sides?
Previously, you learned how to use algebra to solve equations with variables on both sides. Another way is to use a system of linear equations.
Solving Equations Algebraically
and Graphically
Work with a partner. Solve each equation using two methods. Method 1 Use an algebraic method. Method 2 Use a graphical method.
Is the solution the same using both methods?
a.^1 — 2 x + 4 = − —^14 x + 1 b.^2 — 3 x + 4 = (^1) — 3 x + 3
c. − —^23 x − 1 = (^) —^13 x − 4 d.^4 — 5 x + (^) —^75 = 3 x − 3
e. − x + 2.5 = 2 x − 0.5 f. − 3 x + 1.5 = x + 1.
Communicate Your AnswerCommunicate Your Answer
3. How can you use a system of linear equations to solve an equation with variables on both sides? 4. Compare the algebraic method and the graphical method for solving a linear equation with variables on both sides. Describe the advantages and disadvantages of each method.
USING TOOLS
STRATEGICALLY
To be proficient in math, you need to consider the available tools, which may include pencil and paper or a graphing calculator, when solving a mathematical problem.
x
y
2
4
6
− 2 2 4 6
− 2
262 Chapter 5 Solving Systems of Linear Equations
5.5 Lesson^ What You Will LearnWhat You Will Learn
Solve linear equations by graphing. Solve absolute value equations by graphing. Use linear equations to solve real-life problems.
Solving Linear Equations by Graphing
You can use a system of linear equations to solve an equation with variables on both sides.
Previous absolute value equation
Core VocabularyCore Vocabullarry
CoreCore ConceptConcept
Solving Linear Equations by Graphing
Step 1 To solve the equation ax + b = cx + d , write two linear equations. ax + b = cx + d
and
Step 2 Graph the system of linear equations. The x -value of the solution of the system of linear equations is the solution of the equation ax + b = cx + d.
y = ax + b^ y^ =^ cx^ +^ d
Solving an Equation by Graphing
Solve − x + 1 = 2 x − 5 by graphing. Check your solution.
SOLUTION
Step 1 Write a system of linear equations using each side of the original equation. − x + 1 = 2 x − 5
Step 2 Graph the system. y = − x + 1 Equation 1 y = 2 x − 5 Equation 2
The graphs intersect at (2, −1).
So, the solution of the equation is x = 2.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
Solve the equation by graphing. Check your solution.
1.^1 — 2 x − 3 = 2 x 2. − 4 + 9 x = − 3 x + 2
Check − x + 1 = 2 x − 5
−(2) + 1 =
y = − x + 1^ y^ =^2 x^ −^5
x
y 1 1
y = − x + 1
y = 2 x − 5
− 1 − 2
− 4
(2, −1)
264 Chapter 5 Solving Systems of Linear Equations
Solving Real-Life Problems
Modeling with Mathematics
Your family needs to rent a car for a week while on vacation. Company A charges $3.25 per mile plus a flat fee of $125 per week. Company B charges $3 per mile plus a fl at fee of $150 per week. After how many miles of travel are the total costs the same at both companies?
SOLUTION
1. Understand the Problem You know the costs of renting a car from two companies. You are asked to determine how many miles of travel will result in the same total costs at both companies. 2. Make a Plan Use a verbal model to write an equation that represents the problem. Then solve the equation by graphing. 3. Solve the Problem Words Company A Company B
Cost per mile ⋅^
Miles + Flat fee
Cost per mile ⋅^ Miles + Flat fee
Variable Let x be the number of miles traveled. Equation 3.25 x + 125 = 3 x + 150
Solve the equation by graphing.
Step 1 Write a system of linear equations using each side of the original equation. 3.25 x + 125 = 3 x + 150
Step 2 Use a graphing calculator to graph the system.
0 0
600
150
Intersection X=100 Y=
ction
y = 3.25 x + 125
y = 3 x + 150
Because the graphs intersect at (100, 450), the solution of the equation is x = 100. So, the total costs are the same after 100 miles.
4. Look Back One way to check your solution is to solve the equation algebraically, as shown.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
5. WHAT IF? Company C charges $3.30 per mile plus a flat fee of $115 per week. After how many miles are the total costs the same at Company A and Company C?
y = 3.25 x + 125^ y^ =^3 x^ +^150
Check 3.25 x + 125 = 3 x + 150 0.25 x + 125 = 150 0.25 x = 25 x = 100
Section 5.5 Solving Equations by Graphing 265
5.5 Exercises Dynamic Solutions available at^ BigIdeasMath.com
Monitoring Progress and Modeling with MathematicsMonitoring Progress and Modeling with Mathematics
In Exercises 3–6, use the graph to solve the equation. Check your solution.
3. − 2 x + 3 = x 4. − 3 = 4 x + 1
x
y
1
3
− 3 1 3
x
y 1 − 2 2
5. − x − 1 = (^1) — 3 x + 3 6. −^3 — 2 x − 2 = − 4 x + 3
x
y
2
4
− 4 − 2 1
x
y 2 4 − 2
− 4
− 6
In Exercises 7−14, solve the equation by graphing. Check your solution. (See Example 1.)
7. x + 4 = − x 8. 4 x = x + 3 9. x + 5 = − 2 x − 4 10. − 2 x + 6 = 5 x − 1 11.^1 — 2 x − 2 = 9 − 5 x 12. − 5 + (^) —^14 x = 3 x + 6 13. 5 x − 7 = 2( x + 1) 14. −6( x + 4) = − 3 x − 6
In Exercises 15−20, solve the equation by graphing. Determine whether the equation has one solution , no solution , or infi nitely many solutions****.
15. 3 x − 1 = − x + 7 16. 5 x − 4 = 5 x + 1 17. −4(2 − x ) = 4 x − 8 18. − 2 x − 3 = 2( x − 2) 19. − x − 5 = − —^13 (3 x + 5) 20.^1 — 2 (8 x + 3) = 4 x + (^3) — 2
In Exercises 21 and 22, use the graphs to solve the equation. Check your solutions.
21. ∣ x − 4 ∣ = ∣ 3 x ∣ y x − 2 2
− 6
x
y
− 2 2
− 4
− 2
22. ∣ 2 x + 4 ∣ = ∣ x − 1 ∣ y x − 6 − 4
− 6
− 4 x
y
− 1 3
4
− 3
In Exercises 23−30, solve the equation by graphing. Check your solutions. (See Example 2.)
23. ∣ 2 x ∣ = ∣ x + 3 ∣ 24. ∣ 2 x − 6 ∣ = ∣ x ∣ 25. ∣ − x + 4 ∣ = ∣ 2 x − 2 ∣ 26. ∣ x + 2 ∣ = ∣ − 3 x + 6 ∣ 27. ∣ x + 1 ∣ = ∣ x − 5 ∣ 28. ∣ 2 x + 5 ∣ = ∣ − 2 x + 1 ∣ 29. ∣ x − 3 ∣ = 2 ∣ x ∣ 30. 4 ∣ x + 2 ∣ = ∣ 2 x + 7 ∣ 1. REASONING The graphs of the equations y = 3 x − 20 and y = − 2 x + 10 intersect at the point (6, −2). Without solving, find the solution of the equation 3 x − 20 = − 2 x + 10. 2. WRITING Explain how to rewrite the absolute value equation ∣ 2 x − 4 ∣ = ∣ − 5 x + 1 ∣ as two systems of linear equations.
Vocabulary and Core Concept CheckVocabulary and Core Concept Check
