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1© 2015 College Board. All rights reserved. SpringBoard Algebra 2, Unit 1

Solving Linear and Literal EquationsEquations can be used to represent many kinds of situations and relationships in the real world. You can write an equation to describe how your cell phone bill depends on your data usage, or how much interest you will pay before you pay off a purchase on a credit card. By learning to solve equations, you can answer important and interesting questions concerning both your everyday life and the world around you. The properties below are used to solve equations.

Properties of EqualityReflexive Property

Symmetric Property

Transitive Property

Addition Property

Subtraction Property

Multiplication Property

Division Property

a a

If a b, then b a.

If a b and b c, then a c.

If a b, then a 1 c b 1 c.

If a b, then a 2 c b 2 c.

If a b, then ac bc.

If a b and c fi 0, then ac

bc .

When you solve an equation with one variable, you find the value of the variable that makes the equation true. Use the properties of equality to isolate the variable and solve the equation. To apply the properties correctly and keep an equation balanced, whatever you do to one side of the equation, you must also do to the other side.

Solve 25 2 2x 4x 2 41.

Step 1: Use the Addition Property of Equality. Add 2x to each side and combine like terms.

Step 2: Use the Addition Property of Equality. Add 41 to each side and simplify.

Step 3: Use the Division Property of Equality. Divide each side by 6 and simplify.

Solution: x 11

25 2 2x 1 2x 4x 2 41 1 2x

25 6x 2 41

25 1 41 6x 2 41 1 41

66 6x

66 4 6 6x 4 6

11 x

EXAMPLE A

2© 2015 College Board. All rights reserved. SpringBoard Algebra 2, Unit 1

When an equation includes a factor outside of parentheses, remember to use the Distributive Property.

Solve 3a 1 8 2(19 2 a).

Step 1: Use the Distributive Property.

Step 2: Use the Addition Property of Equality. Add 2a to each side and combine like terms.

Step 3: Use the Subtraction Property of Equality. Subtract 8 from each side and simplify.

Step 4: Use the Division Property of Equality. Divide each side by 5 and simplify.

Solution: a 6

3a 1 8 2(19 2 a) 3a 1 8 38 2 2a3a 1 8 1 2a 38 2 2a 1 2a 5a 1 8 38

5a 1 8 2 8 38 2 8 5a 30

5a 4 5 30 4 5 a 6

EXAMPLE B

Sometimes there is more than one way to solve an equation. In fact, adding a step while solving can make your work easier. For instance, you can apply the Multiplication Property of Equality to clear fractions from equations before you work to isolate the variable.

Solve 23

s 13

(s 1 6).

Step 1: Use the Multiplication Property of Equality. Multiply each side by the least common denominator 3 and simplify.

Step 2: Use the Subtraction Property of Equality. Subtract s from each side and combine like terms.

Solution: s 6

323

? s 313

? (s 1 6)

2s s 1 6

2s 2 s s 1 6 2 s

s 6

EXAMPLE C

You can also solve an equation with more than one variable for a particular variable. In this case, you solve for a variable in terms of the other variable or variables. This is called solving a literal equation.Literal equations include two or more variables. Generally you need to solve the equation for one of the variables by isolating it on one side with all of the other variables on the other side. Use the same skills you normally apply to linear equations to isolate the requested variable.

Solving Linear and Literal Equations (continued)

3© 2015 College Board. All rights reserved. SpringBoard Algebra 2, Unit 1

Solve the equation for l: 2l 1 2w 18.

Step 1: Use the Subtraction Property of Equality. Subtract 2w from each side and simplify.

Step 2: Use the Division Property of Equality. Divide each side by 2 and simplify.

Solution: l 9 2 w

2l 1 2w 2 2w 18 2 2w 2l 18 2 2w

l22

w18 2

22

l22

182

2 w22

l 9 2 w

EXAMPLE D

The solution to an equation with two variables is a set of ordered pairs. An ordered pair is a solution to an equation when the values of the ordered pair are substituted for the variables and the result is a true equation.

Is (6, 7) a solution to the equation 2x 2 y 5?

Step 1: Substitute 6 for x and 7 for y.

Step 2: Simplify and see if the equation is true.

Solution: 5 5 is true, so (6, 7) is a solution.

2(6) 2 7 5

12 2 7 5 5 5

?

?

EXAMPLE E

Solving Linear and Literal Equations (continued)

4© 2015 College Board. All rights reserved. SpringBoard Algebra 2, Unit 1

PRACTICESolve each equation. 1. 4x 1 11 14 1 15x 2. 3(d 2 19) 2(29 2 d) 3. 2(8y 2 7) 1 y 5y 2 2

4. 14 (3t 1 27)

34

(3t 1 15) 5. 89

z 1 2 13

(4z 2 6) 6. 2140c 7(6c 1 13)

Solve for each given variable.

7. V lwh, for h 8. V πr2h, for h 9. A 12 bh, for b

Solve for x.

10. x4 1 9y 3 11.

35

(x + 2) y 12. 3x 1 xy 8

Determine whether the given ordered pair is a solution to the equation. 13. 5x 2 3y 28; (5, 21) 14. 2y 2 7x 39; (5, 2) 15. y 3x 2 16; (11, 9)

16. Damon measured the circular pool in his backyard so he could order a cover for it. He found that it had a radius of 8 ft and a circumference of 16 ft. Did he make a mistake in measuring? You may recall that the formula for the circumference of a circle is C 2πr, where C is the circumference and r is the radius, and that the formula for the area of a circle is A πr2. Use mathematics to justify your answer.

Solving Linear and Literal Equations (continued)