Using Formulas to Solve Equations (2)
Example: Find the interest on a savings account with a balance of $2,400 when the interest rate is 3% for 3 years.
Use the formula I = prt.
I = interest earned
p = amount of money invested
r = interest rate
t = time invested
I = prt
Substitute what you know into the formula. I = $2,400 × 3% × 3 years
I = 2,400 × .03 × 3
Simplify the equation. I = $216
Example: How long would you need to invest $3,000 with an interest rate of 3.5% to earn $630?
I = prt
Substitute what you know into the formula. $630 = $3,000 × 3.5% × t
630 = 3,000 × .035t
Simplify the equation. 630 = 105t
Divide both sides of the equation by 105.
61
30
05
=
1
1
0
0
5
5
t
Simplify both sides of the equation. 6 = t
It would take 6 years to earn $630.
Taken From:Algebra success in 20 minites aday
Using Formulas to Solve Equations (1)
LESSON SUMMARY
Formulas solve many real-life problems. In this lesson, you will find out what formulas are and how to use them to solve a variety of problems.
Formulas are special equations that show relationships between quantities. For example, you have probably worked with the formula A=lw. This formula tells you how to find the area of a rectangle.
It tells you to multiply the length times the width to find the area. The formula D = rt tells you to multiply the rate times the time in order to find the distance traveled. When you substitute the information you know into a formula, you can use that to find the information you don’t know. For example, if you travel 55 mph for 3 hours, how far would you travel? Substitute what you know into the equation. Then solve the equation for the variable you don’t know.
Substitute what you know into the formula. D = rt
Multiply 55 times 3. D = 55 · 3
You would travel 165 miles. D = 165
What if you wanted to know how long it would take to travel 300 miles if you were traveling at a speed of 60 mph? All you have to do is substitute what you know into the formula, and then solve for the variable you don’t know.
Substitute what you know into the formula. 300 = 60t
Divide both sides of the equation by 60.
3
6
0
0
0
=
6
6
0
0
t
Simplify both sides of the equation. 5 = t
It would take you 5 hours to travel 300 miles.
This technique works for any formula even though the formula may be very complex.
Taken From:Algebra success in 20 minites aday
Applications
Use a variable to write an equation that represents the problem. Then solve the equation to get the answer to the problem.
Example: Your dental insurance company will pay 80% of all fillings after you have met the deductible of $200. For a recent fillings bill, the insurance company paid $176.What was the original dental bill?
Let x = the original dental bill.
80%(x – $200) = $176
.80(x – 200) = 176
Use the distributive property. .80x – 160 = 176 Add 160 to both sides of the equation. .80x – 160 + 160 = 176 + 160 Simplify both sides of the equation. .80x = 336 Divide both sides of the equation by .80.
.
.
8
8
0
0
x
=
3
.8
3
0
6
Simplify both sides of the equation. x = 420
Example: A trip of 300 miles normally takes you 5 hours when you travel 60 miles an hour. How much faster do you need to drive if you want to make the trip in 4 hours? Let x = how much faster you need to drive. 4(60 + x) = 300 Use the distributive property. 240 + 4x = 300 Subtract 240 from both sides of the equation. 240 – 240 + 4x = 300 – 240 Simplify both sides of the equation. 4x = 60 Divide both sides of the equation by 4.
4
4
x
=
6
4
0
36. A corporation pays 1 2 of the health insurance premium and an additional $200 a month toward other benefits such as dental insurance. Your total benefit package is $450 a month.How much is your health insurance premium each month?
37. A credit card states that your payment will be a minimum of $15 plus 1% of your unpaid balance. Your unpaid balance is $2,365.What is your payment this month?
38. The length of a room is 3 more than twice the width of the room. The perimeter of the room is 66 feet. What are the dimensions of the room? (Let x = the width of the room.)
39. Suppose a roast should be cooked for 45 minutes plus 10 more minutes for every pound the roast weighs. If a roast is properly cooked in 3 hours, how much did it weigh?
40. If you work Christmas Day, your company will pay you double time plus a $50 bonus. If you earn $218 for an 8-hour day, what is your hourly rate of pay?
Skill Building until Next Time
If you have dental work done, what is your deductible? What percent coverage does your insurance provide for preventive care, such as having your teeth cleaned? Is there a difference in coverage when you have a filling, root canal, or crown? If you don’t have dental insurance, call an insurance company and inquire about the different types of coverage. What would it cost you to have a cleaning and a root canal? Set up an equation using the information you obtain, and then solve it.
Taken From:Algebra success in 20 minites aday
Equations Without a Variable in the Answer (2)
Subtract 3 from both sides of the equation. 5x + 3 – 3 = 5x + 3 – 3
Simplify both sides of the equation. 5x = 5x
Subtract 5x from both sides of the equation. 5x – 5x = 5x – 5x
Simplify both sides of the equation. 0 = 0
This is called an identity because when you simplified the equation, the left side of the equation equaled the right side.When this happens, any number can be a solution. It doesn’t matter what value you use to replace the variable, the left side of the equation will always equal the right side of the equation. Because any real number can be a solution, you have an infinite (endless) number of solutions. The mathematical notation for any real number is the capital letter R. You use this R to represent the answer to this equation.
Practice
______31. 2x + 5 = 2(x + 5)
______32. 5x + 3x + 2 = 4(2x + 1) – 2
______33. 4x + 3(x + 3) = 7(x + 1) + 2
______34. 3x – 6 = 3(1 + 2x)
______35. 11x – 3 + 4x = 3(5x – 2) + 4
Taken From:Algebra success in 20 minites aday
Equations Without a Variable in the Answer (1)
In some equations, all the variables will be eliminated. These are special cases, and the equations may or may not have solutions.
Case 1 3x + 15 = 3(x + 10)
Use the distributive property. 3x + 15 = 3x + 30
Subtract 15 from both sides of the equation. 3x + 15 – 15 = 3x + 30 – 15
Simplify both sides of the equation. 3x = 3x + 15
Subtract 3x from both sides of the equation. 3x – 3x = 3x – 3x + 15
Simplify both sides of the equation. 0 = 15
Your common sense tells you that 0 will never equal 15. This means there is no value of the variable that will make the equation true because 0 will never equal 15. Since there is no value of x that will ever make the equation true, there is no solution.When there is no solution, it is called an empty set. This notation is used for the empty set.
Case 2 5x + 3 = 5(x – 1) + 8
Use the distributive property. 5x + 3 = 5x – 5 + 8
Combine similar terms. 5x + 3 = 5x + 3
Taken From:Algebra success in 20 minites aday
Solving More Complex Equations
In this section, you will use the distributive property and combine similar terms to simplify equations. The equations will also contain variables on both sides of the equation. Look at the following examples very carefully.
Example: 2x + 3 + 3x = 4x – 7 + 9
Combine similar terms on both sides of the equation. 5x + 3 = 4x + 2
Subtract 3 from both sides of the equation. 5x + 3 – 3 = 4x + 2 – 3
Simplify both sides of the equation. 5x = 4x – 1
Subtract 4x from both sides of the equation. 5x – 4x = 4x – 4x – 1
Simplify both sides of the equation. x = –1
Example: 5x + 3 – 2x = 2(x – 3) + 5
Use the distributive property. 5x + 3 – 2x = 2x – 6 + 5
Combine similar terms on both sides of the equation. 3x + 3 = 2x – 1
Subtract 3 from both sides of the equation. 3x + 3 – 3 = 2x – 1 – 3
Simplify both sides of the equation. 3x = 2x – 4
Subtract 2x from both sides of the equation. 3x – 2x = 2x – 2x – 4
Simplify both sides of the equation. x = –4
Practice
Write out all your steps.
______21. 4x – 3 + 3x = –2x + 5 – 17
______22. 12x + 2 – 5x = 3(x + 5) + 3
______23. 12x – 11x + 5 = –2(x + 3) + 5
______24. 5 – (x – 3) = 10 – 3x
______25. 3x + 2 – 6x = 8 + 3(x + 2)
______26. 9x – (x + 3) = 2(x + 4) + 7
______27. 5(2x + 3) = 4(x – 1) + 1
______28. 1.2x + 2(x + .6) = 5x – 9.8 – 4x
______29. 13x – 2(3x – 4) = 6(x + 2) – 4
______30. 13x – 3(–2x – 3) = 4(x + 4) – 12
Taken From:Algebra success in 20 minites aday
Using the Distributive Property
To solve some equations with variables on both sides of the equation, you will need to use the distributive property that you learned about in Lesson 3. The distributive property tells you to multiply the number outside the parentheses by each term inside the parentheses in equations like 2(x + 2) = 2x + 4 and 3(a – b) = 3a – 3b
Example: 5x + 3 = 3(x + 5)
Use the distributive property. 5x + 3 = 3x + 15
Subtract 3 from both sides of the equation. 5x + 3 – 3 = 3x + 15 – 3
Simplify both sides of the equation. 5x = 3x + 12
Subtract 3x from both sides of the equation. 5x – 3x = 3x – 3x + 12
Simplify both sides of the equation. 2x = 12
Divide both sides of the equation by 2.
2
2
x
=
1
2
2
Simplify both sides of the equation. x = 6
Example: 4x + 6 = –2(3x + 4)
Use the distributive property. 4x + 6 = –6x – 8
Subtract 6 from both sides of the equation. 4x + 6 – 6 = –6x – 8 – 6
Simplify both sides of the equation. 4x = –6x – 14
Add 6x to both sides of the equation. 4x + 6x = –6x + 6x – 14
Simplify both sides of the equation. 10x = –14
Divide both sides of the equation by 10.
1
1
0
0
x
= –
11
40
Simplify both sides of the equation.
Practice
Be sure to write out your steps!
______11. 4x + 2 = 2(x + 3)
______12. 3x + 3 = 2(x – 3)
______13. 4x + 6 = 2(3x – 4)
______14. 6x – 2 = 2(2x + 3)
______15. 6x – 3 = 3(3x – 5)
______16. 4.2x + 6 = 2(2x + 3)
______17. 11x – 5 = 7(x – 2)
______18. 8x + 5 = –3(x + 2)
______19. 4x + 12 = 3(4 – x)
______20. 11x – 3 = –3(–x + 3)
Taken From:Algebra success in 20 minites aday
Solving Equations with Variables on Both Sides of the Equation (2)
Example: 3x – 2 = 2x + 5
Add 2 to both sides of the equation. 3x – 2 + 2 = 2x + 5 + 2
Simplify both sides of the equation. 3x = 2x + 7
Subtract 2x from both sides of the equation. 3x – 2x = 2x – 2x + 7
Simplify both sides of the equation. x = 7
Example: 6x – 5 = 3x + 4
Add 5 to both sides of the equation. 6x – 5 + 5 = 3x + 4 + 5
Simplify both sides of the equation. 6x = 3x + 9
Subtract 3x from both sides of the equation. 6x – 3x = 3x – 3x + 9
Simplify both sides of the equation. 3x = 9
Divide both sides of the equation by 3.
3
3
x
=
93
Simplify both sides of the equation. x = 3
Practice
You can see that you are starting to work more complex equations. Be sure to write out all steps. This may seem like unnecessary work, but taking the time to write out all the steps actually saves you time. You will be less apt to make a mistake. Also, it is easier to find and correct a mistake if you write out your steps.
______ 1. 7x + 2 = 3x – 6
______ 2. 5x – 3 = 3x + 7
______ 3. 8x + 2 = 4x – 6
______ 4. 9x + 4 = 8x + 12
______ 5. 9x + 12 = 6x + 3
______ 6. 6x + 4 = 2x + 12
______ 7. 8x + 7 = 2x + 3
______ 8. 4x – 5 = 5x + 1
______ 9. 2x + 5 = 5x – 1
______10. 3.5x + 1 = 3x + 4
Taken From:Algebra success in 20 minites aday
Solving Equations with Variables on Both Sides of the Equation (1)
LESSON SUMMARY
You know that when you solve equations, the object is to get the variable on a side by itself or, in other words, isolate the variable. In this lesson, you will solve equations that have variables on both sides of the equation.
What Do You Do When You Have Variables on
Both Sides of the Equation?
To solve the equation, 2x + 5 = x – 3, you can start by using the techniques you already know. You know that you want to isolate the variable on a side by itself. You don’t want the 5 with the variable 2x. Get rid of the 5 by subtracting 5 from both sides of the equation.
Example: 2x + 5 = x – 3
Subtract 5 from both sides of the equation. 2x + 5 – 5 = x – 3 – 5
Simplify both sides of the equation. 2x = x – 8
You still don’t have the variable isolated because you have variables on both sides of the equation. You need to get rid of the x on the right side of the equation, and you know that x – x = 0, so subtract x from both sides of the equation.
Subtract x from both sides of the equation. 2x – x = x – x – 8
If there is no number (coefficient) in front of the x, it is understood to be 1. So 2x minus 1x is 1x or x. And 1x minus 1x is zero.
Simplify both sides of the equation to get your answer. x = –8
Taken From:Algebra success in 20 minites aday
Solving Equations That Have a Fraction in Front of the Variable
How do you eliminate the number in front of the variable (coefficient) when it is a fraction? There are many approaches you could use to solve an equation with a fraction for a coefficient, but let’s use the shortest method. The shortest method is to multiply both sides of the equation by something called the multiplicative inverse of the coefficient. You’re probably wondering, “What is the multiplicative inverse of a number?”Take a look at these examples:
The multiplicative inverse of 2, which can be written as 21 , is
12 . The multiplicative inverse of 5, which can be written as 51
, is 15 . The multiplicative inverse of 23 is 32 . The multiplicative inverse of – 34 is – 43 . You probably get the
idea now—you get the multiplicative inverse of a number by inverting the number. In other words, by turning it upside down A number times its multiplicative inverse will always equal 1. Thus, 45 ·
54 = 1. Here’s how you use the multiplicative inverse to solve equations with a fraction in front of the variable:
Example:
23
x + 1 = 5
Subtract 1 from both sides of the equation.
23
x + 1 – 1 = 5 – 1
Simplify both sides of the equation.
23
x = 4
Multiply both sides of the equation by the multiplicative inverse
of the coefficient.
23
x ·
32
= 4 ·
32
Simplify both sides of the equation. x = 6
Example:
35
x – 2 = 7
Add 2 to both sides of the equation.
35
x – 2 + 2 = 7 + 2
Simplify both sides of the equation.
35
x = 9
Multiply both sides of the equation by
53
.
35
x ·
53
= 9 ·
53
Simplify both sides of the equation.
Practice
Solve the equations. Show all the steps needed to solve each equation.
______26. 47 x – 1 = 11
______27. 56 x + 4 = 14
______28. – 34 x + 5 = 2
______29. 7 = 1 2 x + 2
______30. 36 = – 43 x + 4
______31. 1 2 7 x – 2 = 15
______32. 58 x + 3 = 3
______33. – 4 5 x – 14 = –2
______34. 65 x + 2 = 7
______35. – 2 3 x – 1 = 2
Skill Building until Next Time
It is recommended that you spend no more than 25% of your income on housing. Calculate the recommended housing expenditure based on your income or your parents’ income. Are you within the guideline?
What other expenses need to be included when calculating one’s housing expense other than the rent or mortgage payment?
Taken From:Algebra success in 20 minites aday