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Chapter 1 Review Advanced Algebra 1

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**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

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**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 –

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**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

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**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

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**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 –

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**Since – is added to z, add to both sides. + 3 4 z = 2**

5 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

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**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.

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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

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**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

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**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

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**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

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**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

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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.

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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

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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

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**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.

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Solve Method 1 Use fraction operations.

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**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.

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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

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**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

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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.

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**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

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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

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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

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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

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**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.

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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.

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**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

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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}.

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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}.

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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}.

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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.

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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

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**The two windows below are similar**

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

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

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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

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