Opening the book…
LEARNING GOALS
By the end of this lesson, you should understand
- that multiplication undoes a division and division undoes a multiplication;
- how to work a simple multiplication or division problem backwards.
By the end of this lesson, you should be able to
- solve one-step linear equations involving a product or a quotient by working the problem backwards,
- solve one-step linear equations by multiplying or dividing both sides of the equation by the same number.
FUNDAMENTALS OF THE LESSON
Reminder: In algebra, we use
- numbers and letters to represent quantities known and unknown,
- expressions to represent operations involving numbers.
TRY THESE
How Much Money Were They Making Before The Changes?
Eddy has been working at the same small company for a long time. His hourly wage has varied.
1. In the year 2000, Eddy received a promotion. His hourly rate tripled, coming to be $24 an hour.
A. How much was Eddy making right before his hourly rate tripled?
B. What operation did you use to get that result? With what numbers?
2. Let x represent Eddy’s hourly rate before his promotion.
A. Using x, how would you write an expression for Eddy’s hourly rate after his promotion? Remember that he was making x, and his rate tripled.
B. How would you write that the amount you wrote in part A is equal to 24?
C. Without carrying out any calculations, how would you write what x is equal to? Remember the operation you did. Do not do the calculation, just indicate it.
D. From your answer to part C, determine what x is equal to.
What you obtained in 2A is a mathematical expression. A mathematical expression is a group of symbols including numbers, variables and operations, used to show the value of something.
An equation is a statement saying that two mathematical expressions are equal to each other. Examples of equations are your answers to 2B, 2C, and 2D.
3. Now you have the equation 2x = 14.
A. Can you find x by performing operations that are similar to the ones in question 2?
B. What operation was originally being done with 2 on the left side of the equation? How did the 2 “move” to the right?
C. Now substitute the value you obtained for x into the equation 2x = 14. What do you observe?
In Question 3 you found the value of x that turned the equation 2x = 14 into an equality which is a true statement. That value is 7. 7 is the solution of the equation 2x = 14 because when you replace x with 7, the equation becomes an equality. Solving an equation means finding its solution.
4. Practice what you just learned and solve the following equations. Make sure you master this method.
A. 5y = 55
B. -4n = 25
C. -5.1 = -6x
5. Sometimes students are taught to “divide both sides” to solve a problem like 2x = 14. In your opinion, what are the advantages and disadvantages of this method compared to what we practiced above?
6. In the year 2009, a recession hit. Eddy’s company had to slash wages and his hourly rate became half of what it was right before the recession. After the cut, Eddy started making $25 per hour.
A. How much was he making right before the cut?
B. What operation did you do in order to find that amount?
C. If you were to denote Eddy’s hourly wage before the cut by x, how would you write the fact that his new hourly rate is $25?
D. Without doing any calculations, how would you write what x is equal to? Remember the operation you did. Do not finish the calculation.
E. From your answer to part C, determine what x is equal to.
7. Now you have the equation x ÷ 4 = 5.
A. Can you find x by performing operations that are similar to the ones in Question 6?
B. What operation was originally being done with 4 on the left side of the equation?
C. How did the 4 “move” to the right?
8. Practice what you just learned and solve the following equations.
A. y ÷ 3.2 = 8.5
B. \(\dfrac{2n}{3} = -10\)
C. \(-\dfrac{x}{6} = -7\)
D. \(187 = \dfrac{11x}{5}\)
FURTHER APPLICATIONS
Solve the following equations that include fractions. Try using the first method you learned today:
9. \(\dfrac{2}{3}x = \dfrac{5}{7}\)
10. \(-\dfrac{3}{4}y = \dfrac{3}{10}\)
11. \(-\dfrac{10}{3}z = -\dfrac{15}{9}\)
Questions: Linear Equations (Part 2)
Solve the following equations using the primary method from this lesson.
- 8x = -33
- -51 = -3x
- 0.25y = 10
- \(\dfrac{y}{8} =\dfrac{3}{2}\)
- \(\dfrac{n}{0.1} = 6.4\)
- \(\dfrac{3n}{10} = -1.2\)
- \(\dfrac{5}{3}y = 7\)
This page titled 5.26: Fundamentals 24 - Linear Equations (Part 2) was last modified on Wed, 15 May 2024 07:37:23 GMT and is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Carnegie Math Pathways (WestEd) .
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