1. Chapter 1 Review
Advanced Algebra 1

2. Give two ways to write each algebraic expression in words.
A r B. q – 3
the sum of 9 and r
the difference of q and 3
9 increased by r
3 less than q
C. 7m D. j  6
the product of m and 7
the quotient of j and 6
m times 7
j divided by 6

3. Solve the equation. Check your answer.
Since 8 is subtracted from y, add 8 to both sides to undo the subtraction.
y = 32
Check
y – 8 = 24
To check your solution, substitute 32 for y in the original equation.
32 – 

4. Solve the equation. Check your answer.
= z – 7 16 5
Since is subtracted from z, add to
both sides to undo the subtraction. 7 16 + 7 16
= z 3 4
Check
= z – 7 16 5
To check your solution, substitute for z in the original equation. 3 4 3 4 5 16 7 – 5 16 

5. Solve the equation. Check your answer.1 2
d = 1
Since is added to d, subtract from both sides to undo the addition. 1 2 – 1 2
d = 1 2
Check
d = 1 1 2
To check your solution, substitute for d in the original equation. 1 2 1 2
1 1 

6. Since – is added to p, add to both sides.
Solve –
+ p = – 2 11 5 + 5 11 5 11
Since – is added to p, add to both sides.
p = 3 11
Check
+ p = – 2 11 5 –
To check your solution, substitute for p in the original equation. 3 11 2 5 11 – 3 + 2 11 – 

7. Since – is added to z, add to both sides. + 3 4 z = 25 4
Solve –
+ z = Check your answer. –
+ z = 5 4 3 3 4
Since – is added to z, add to both sides. + 3 4
z = 2
Check
+ z = 5 4 3 –
To check your solution, substitute 2 for z in the original equation. – 5 3 4
+ 2 5 4 

8. decrease in population
Over 20 years, the population of a town decreased by 275 people to a population of 850. Write and solve an equation to find the original population.
original population
minus
current population
decrease in population is
p – d = c
p – 275 =850
Substitute 275 for d and 850 for c.
p – 275 = 850
Since 275 is subtracted from p, add 275 to both sides to undo the subtraction.
p =1125
The original population was 1125 people.

9. Solve the equation.
–8 = j 3
Since j is divided by 3, multiply both sides by 3 to undo the division.
–24 = j
Check
–8 = j 3 –8
–24 3
To check your solution, substitute –24 for j in the original equation.
–8 –8 

10. Solve the equation. Check your answer.
–13 = y 3
Since y is divided by 3, multiply both sides by 3 to undo the division.
–39 = y y
Check
–13 = 3
–13
–39 3
To check your solution, substitute –39 for y in the original equation.
–13 –13 

11. Solve the equation. Check your answer.
Since y is multiplied by 9, divide both sides by 9 to undo the multiplication.
y = 12
Check 9y = 108
To check your solution, substitute 12 for y in the original equation.
9(12) 108 

12. To check your solution, substitute 24 for w in the original equation.
Solve the equation. 5
w = 20 6
The reciprocal of is Since w is multiplied by , multiply both sides by 5 6
w = 24
Check
w = 20 5 6
To check your solution, substitute 24 for w in the original equation.
20
20 20 

13. Solve the equation. 4j 2 = 6 3 is the same as j.
The reciprocal of is . Since j is multiplied by , multiply both sides by . 4 6
j = 1

14. Ciro puts of the money he earns from mowing lawns into a college education fund. This year Ciro added $285 to his college education fund. Write and solve an equation to find how much money Ciro earned mowing lawns this year. 1 4
one-fourth times earnings equals college fund
Write an equation to represent the relationship.
Substitute 285 for c. Since m is divided by 4, multiply both sides by 4 to undo the division.
m = $1140
Ciro earned $1140 mowing lawns.

15. Solve 18 = 4a + 10.
18 = 4a + 10
First a is multiplied by 4. Then 10 is added. Work backward: Subtract 10 from both sides.
– – 10
8 = 4a
Since a is multiplied by 4, divide both sides by 4 to undo the multiplication.
8 = 4a 4
2 = a

16. Solve –4 + 7x = 3.
–4 + 7x = 3
First x is multiplied by 7. Then –4 is added. Work backward: Add 4 to both sides.
7x = 7
Since x is multiplied by 7, divide both sides by 7 to undo the multiplication.
7x = 7 7
x = 1

17. Method 1 Use fraction operations.
Solve
Method 1 Use fraction operations.
Since is subtracted from , add to both sides to undo the subtraction. 3 4 y 8
Since y is divided by 8, multiply both sides by 8 to undo the division.

18. Solve
Method 1 Use fraction operations.

19. Method 1 Use fraction operations.
Solve
Method 1 Use fraction operations.
Since is subtracted from , add to both sides to undo the subtraction. 1 3 n 5
Simplify.

20. Solve 8x – x = –15.
8x – 21 – 5x = –15
8x – 5x – 21 = –15
Use the Commutative Property of Addition.
3x – 21 = –15
Combine like terms.
Since 21 is subtracted from 3x, add 21 to both sides to undo the subtraction.
3x = 6
Since x is multiplied by 3, divide both sides by 3 to undo the multiplication.
x = 2

21. Solve 10y – (4y + 8) = –20 10y + (–1)(4y + 8) = –20
Write subtraction as addition of the opposite.
10y + (–1)(4y + 8) = –20
10y + (–1)(4y) + (–1)( 8) = –20
Distribute –1 on the left side.
10y – 4y – 8 = –20
Simplify.
6y – 8 = –20
Combine like terms.
Since 8 is subtracted from 6y, add 8 to both sides to undo the subtraction.
6y = –12
6y = –12
Since y is multiplied by 6, divide both sides by 6 to undo the multiplication.
y = –2

22. Solve 2a + 3 – 8a = 8.
2a + 3 – 8a = 8
2a – 8a + 3 = 8
Use the Commutative Property of Addition.
–6a + 3 = 8
Combine like terms.
– 3 – 3
Since 3 is added to –6a, subtract 3 from both sides to undo the addition.
–6a = 5
Since a is multiplied by –6, divide both sides by –6 to undo the multiplication.

23. Solve –2(3 – d) = 4 –2(3 – d) = 4 (–2)(3) + (–2)(–d) = 4 –6 + 2d = 4
Distribute –2 on the left side.
–6 + 2d = 4
Simplify.
–6 + 2d = 4
Add 6 to both sides.
2d = 10
2d = 10
Since d is multiplied by 2, divide both sides by 2 to undo the multiplication.
d = 5

24. Solve 7n – 2 = 5n + 6.
7n – 2 = 5n + 6
To collect the variable terms on one side, subtract 5n from both sides.
–5n –5n
2n – 2 =
2n =
Since n is multiplied by 2, divide both sides by 2 to undo the multiplication.
n = 4

25. Solve 4b + 2 = 3b.
4b + 2 = 3b
To collect the variable terms on one side, subtract 3b from both sides.
–3b –3b
b + 2 = 0
– 2 – 2
b = –2

26. Solve
Distribute to the expression in parentheses. 1 2
To collect the variable terms on one side, subtract b from both sides. 1 2
3 = b – 1
Since 1 is subtracted from b, add 1 to both sides.
4 = b

27. The formula for the area of a triangle is A = bh,
where b is the length of the base, and h is the height. Solve for h.
A = bh
Locate h in the equation.
Since bh is multiplied by , divide both sides by to undo the multiplication.
2A = bh
Since h is multiplied by b, divide both sides by b to undo the multiplication.

28. A. Solve x + y = 15 for x.
x + y = 15
Locate x in the equation.
–y –y
x = –y + 15
Since y is added to x, subtract y from both sides to undo the addition.
B. Solve pq = x for q.
pq = x
Locate q in the equation.
Since q is multiplied by p, divide both sides by p to undo the multiplication.

29. Solve for the indicated variable.1. 2.
3. 2x + 7y = 14 for y 4.
for h
P = R – C for C
C = R – P
for m
m = x(k – 6 ) 5.
for C
C = Rt + S

30. Solve the equation.
3|x + 7| = 24
Since |x + 7| is multiplied by 3, divide both sides by 3 to undo the multiplication.
Think: What numbers are 8 units from 0?
|x + 7| = 8
Case 1
x + 7 = 8
Case 2
x + 7 = –8
– 7 –7
– 7 – 7
x = 1
x = –15
Rewrite the equations as two cases. Since 7 is added to x subtract 7 from both sides of each equation.
The solutions are {1, –15}.

31. Solve the equation.
|x| – 3 = 4
Since 3 is subtracted from |x|, add 3 to both sides.
|x| – 3 = 4
|x| = 7
Think: What numbers are 7 units from 0?
Case 1
x = 7
Case 2
x = –7
Rewrite the equation as two cases.
The solutions are {7, –7}.

32. Solve the equation.
8 = |x + 2|  8
Since 8 is subtracted from |x + 2|, add 8 to both sides to undo the subtraction.
8 = |x + 2|  8
0 = |x + 2|
There is only one case. Since 2 is added to x, subtract 2 from both sides to undo the addition.
0 = x + 2
 2
2 = x
The solution is {2}.

33. Solve the equation.
2  |2x  5| = 7
Since 2 is added to –|2x – 5|, subtract 2 from both sides to undo the addition.
2  |2x  5| = 7
 2
 |2x  5| = 5
Since |2x – 5| is multiplied by negative 1, divide both sides by negative 1.
|2x  5| = 5
Absolute value cannot be negative.
This equation has no solution.

34. Just as we solved for variables in earlier proportions, we can solve for variables to find unknown sides in similar figures. Set up the corresponding sides as a proportion and then solve for x.
Ratios
x/12
and
5/10 x
10x = 60
x = 6

35. The two windows below are similar
The two windows below are similar. Find the unknown width of the larger window.

36. These two buildings are similar. Find the height of the large building.

37. Ex: The dosage of a certain medication is 2 mg for every 80 lbs of body weight. How many milligrams of this medication are required for a person who weighs 220 lbs?
Use this rate to determine the dosage for 220-lbs by setting up a proportion (match units) 
Let x = required dosage
x mg =
 2(220) = 80x
220 lbs
 440 = 80x
 x = 5.5 mg