Download Practice A Equations

Practice A
Equations
Unit 1
Getting
Ready
Equations 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 usage time, or how much interest you will pay if you take a
certain amount of time to 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.
Review the properties below, which are used to solve equations.
Properties of Equality
Reflexive Property
a=a
Symmetric Property
If a = b, then b = a.
Transitive Property
If a = b and b = c, then a = c.
Addition Property
If a = b, then a + c = b + c.
Subtraction Property
If a = b, then a − c = b − c.
Multiplication Property
If a = b, then ac = bc.
Division Property
If a = b and c ≠ 0, then c__
​ a ​ = ​c__
 b ​ .
© 2011 College Board. All rights reserved.
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. Remember that 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.
EXAMPLE 1
Solve 25 − 2x = 4x − 41.
Step 1:
Step 2:
Step 3:
Use the Addition Property of Equality. Add 2x to each side
and combine like terms.
Use the Addition Property of Equality. Add 41 to each side
and simplify.
Use the Division Property of Equality. Divide each side by
6 and simplify.
25 − 2x + 2x = 4x − 41 + 2x
25 = 6x − 41
25 ­+ 41 = 6x − 41 + 41
66 = 6x
66 ÷ 6 = 6x ÷ 6
11 = x
Solution: x = 11
When an equation includes a factor outside of parentheses, remember to use
the distributive property.
Algebra 2, Unit 1 • Linear Systems and Matrices A-1
Unit 1
Practice A
Getting
Ready
Equations continued
EXAMPLE 2
Solve 3a + 8 = 2(19 − a).
Step 1:
Use the distributive property.
Step 2:
Use the Addition Property of Equality. Add 2a to each side
and combine like terms.
Use the Subtraction Property of Equality. Subtract 8 from each
side and simplify.
Use the Division Property of Equality. Divide each side by
5 and simplify.
Step 3:
Step 4:
3a + 8 = 2 ∙ 19 − 2a
3a + 8 = 38 − 2a
3a + 8 + 2a = 38 − 2a + 2a
5a + 8 = 38
5a + 8 − 8 = 38 − 8
5a = 30
5a ÷ 5 = 30 ÷ 5
a = 6
Solution: a = 6
There is sometimes more than one way to solve an equation. In fact, sometimes
adding a step while solving can actually 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.
Step 1:
Use the Multiplication Property of Equality. Multiply each
side by 3 and simplify.
Step 2:
Use the Subtraction Property of Equality. Subtract s from
each side and combine like terms.
​ 1 ​ (s + 6)
3 ∙ __
​ 2 ​ s = 3 ∙ __
3
3
2s = s + 6
2s − s = s + 6 − s
s=6
Solution: s = 6
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.
EXAMPLE 4
Solve the equation for l.
2l + 2w = 18
Step 1:
Use the Subtraction Property of Equality. Subtract 2w
from each side and simplify.
2l + 2w − 2w = 18 − 2w
2l = 18 − 2w
Step 2:
Use the Division Property of Equality. Divide each side
by 2 and simplify. 
2w 
 
2l ÷ 2 = _______
​ 18 − ​
2
2l ÷ 2 = ___
​ 18 ​ − ___
​ 2w ​ 
2
2
l = 9 − w
Solution: l = 9 – w
A-2 Getting Ready Practice A Equations
© 2011 College Board. All rights reserved.
EXAMPLE 3
​ 1 ​  (s + 6).
Solve __
​ 2 ​  s = __
3
3
Unit 1
Practice A
Getting
Ready
Equations continued
The solution to an equation with two variables is a set of ordered pairs. An
ordered pair is a solution to an equation if you substitute the values of the
variables and the result is a true equation.
EXAMPLE 5
Is (6, 7) a solution to the equation 2x − y = 5?
Step 1:
Step 2:
2(6) − 7 ≟ 5
12 − 7 ≟ 5
5 = 5
Substitute 6 for x and 7 for y.
Simplify and see if the equation is true.
Solution: 5 = 5 is true, so (6, 7) is a solution.
TRY THESE
Solve each equation.
1. 4x + 11 = 14 + 5x
2. 3(d − 19) = 2(29 −d)
3. 2(8y − 7) + y = 5y − 2
1  ​(3t + 27) = __
​ 3 ​ (3t + 15)
4. ​ __
4
4
5. __
​ 8 ​ z + 2 = __
​ 1 ​ (4z − 6)
9
3
6. −140c = 7(6c + 13)
8. V = πr2h, for h
1  ​bh, for b
9. A = ​ __
2
Solve for each given variable.
7. V = lwh, for h
Solve for x.
10. __
​ x  ​+ 9y = 3
4
11. __
​ 3 ​ (x + 2) = y
5
12. 3x + xy = 8
© 2011 College Board. All rights reserved.
Determine whether the given ordered pair is a solution to the equation.
13. 5x − 3y = 28; (5, −1)
14. 2y − 7x = 39; (5, 2)
15. y = 3x − 16; (11, 9)
16. Damon measured the circular pool in his back yard so he could order a
cover for it. He found that it had a radius of 8 ft and a circumference of
16 ft. Can you tell whether he made a mistake in measuring? You may
recall that he formula for circumference of a circle is C = 2πr, where C is
circumference and r is the radius and that the formula for area of a circle
is A = πr2. Explain using mathematics to justify your answer.
Algebra 2, Unit 1 • Linear Systems and Matrices A-3