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REVISION WITH ANSWERS – PLEASE REVISE

Algebra 1 Lesson 1.1 Variables and Expressions

Question 1:

Determine whichexpression containsone∨more variables .

A. -7

B. 5y2

C. π

D. 64

Question 2

Write an algebraic expression for the verbal expression.

Three added ¿twice anumber

Solution :2x+3∨3+2x

Question 3

the|expression

34

Solution :34=3×3×3×3=81

Question 4

Write averbal expression for thealgebraic expression

3 s2−5 t

Solution :The product of 5∧t , subtracted ¿ the product of 3∧the squareof s .

Mathematics department 2012/20131

Question 5:

Acar rental company chargesdaily a flat rate of 100dirhams∧an additional7dirhams

per kilometer k .Write analgrebraic expressionthat represents the totaldaily cost of renting

acar ¿ this company .

Solution :7k+100∨100+7k

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Algebra 1 lesson 1-2: Solving Equations by Adding or Subtracting

Question 1:

Determine if the given valueis a solution¿ the equation

m−5=−8

A. 5

B. 3

C. -3

D. -5

Question 2:

Solvethe equation .Then ˇyour solution.

11+x=13

Solution :

11+x=13

11+x+(−11)=13+(−11)

x=2

:̌

11+x=13

11+2=13

13=13✓

The solutionis2.


Question 3:

If y−9=16 ,what is the value of y−6?

Solution :

y−9=3

y−9+9=3+9

y=12

So , y−6=12−6

y−6=6

Question 4:


x− 53=−1

2

Solution :

x−53+ 5

3=−1

2+ 5

3

x=−12

+ 53=−3

6+10

6∨7

6

So ,x=76

:̌

76−5

3=−1

2

76−10

6=−1

2

−36

=−12

−12

=−12✓

The solutionis 76

Question 5:


Write an equation for the problem. Then, solve the equation and check the solution.

The∑ of anumber∧−13is−28.What isthe number ?

Solution

Let x represent the number .

x+(−13 )=−28

x+(−13 )+13=−28+13

x=−15

:̌

−15+(−13 )=−28

−28=−28✓

The solutionis−15.

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Algebra 1 lesson 1-3: Solving Equations by Multiplying or Dividing

Question 1:

If 8 t=20 ,what isthe value of 16 t ?

A. 40

B. 30

C. 20

D. 10

Question 2:



35f=−18

Solution :

35f=−18

35f ×( 5

3 )=−18×( 53 )

f=−30

:̌

35f=−18

35(−30)=−18

−18=−18✓

The solutionis−30.

Question 3:


0.36 x=9.135

Solution :

0.36 x0.36

=9.1350.36

x=2038

:̌

0.36 x=9.135

0.36×( 2038 )=9.135

9.135=9.135✓

Question 4:



− y5

=−35

Solution :

− y5×(−5)=−35×(−5)

y=175

:̌

− y5

=−35

−1755

=−35

−35=−35✓

The solutionis175

Question 5:

¿1993 , theworldrecord for thelargest ¿doughnut was established .Thedoughnut weighed1.5 tons∧¿hadacircumference of 50 feet .Using

the formula for the circumference of the C̊=πd , find the diameterof that doughnut .

Solution

C=πd

50=πd

50π

=πdπ

d=50πfeet

:̌

C=πd

50=π × 50π


50=50✓

So ,the diameter was 50πfeet .

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Algebra 1 lesson 1-4: Solving Two-Step and Multi-Step Equations

Question 1:

Determineif the given valueis a solution¿ the given equation

−4 y−9=3

A. -3

B. -4

C. 3

D. 4

Question 2:


−10 x+2=−18

Solution :

−10 x+2=−18

−10 x+2−2=−18−2

−10 x=−20

−10x−10

=−20−10

x=2

:̌

−10 x+2=−18

−10(2)+2=−18

−20+2=−18


−18=−18✓

The solutionis2.

Question 3:


−17=4 ( p−5)

Solution :

−17=4 ( p−5)

−17=4 p−20

−17+20=4 p−20+20

3=4 p

34=4 p

4

p=34∨p=0.75

:̌

−17=4 ( p−5)

−17=4∗( 34−5)

−17=4∗(−174 )

−17=−17✓

The solutionis 34.

Question 4:


The sum of 3 times a number and 5 is 17. Find the number.

Solution :

Let x be the number .

3 x+5=17

3 x+5−5=17−5

3 x=12

3x3

=123

x=4

:̌

3(4)+5=17

12+5=17

17=17✓

The solutionis 4.So , the number is4.

Question 5:

Fatimawants¿buy adress that costs600dirhams . Shehas150dirhams∈her savings

¿ shereceives anallowance of 75dirhamseveryweek .Set up anequation¿determine

the number of weeks she needs¿ save her allowancemoney∈order ¿haveenough

money for the dress .

Solution :

Let x be the number of weeks Fatimaneeds ¿ save her allowancemoney .

150+75 x=600

150+75 x−150=600−150

150+75 x=600

75 x=450

75x75

=45075


x=6

:̌

150+75 x=600

150+75(6)=600

600=600✓

The solutionis6. So ,Fatimaneeds ¿ save her allowancemoney for 6weeks .

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Algebra 1 Lesson 1-5 Solving Equations with Variables on Both Sides

Question 1:

Determine if the given valueis a solution¿ the given equation

8 t=4 t+16

A. -4

B. -3

C. 3

D. 4

Question 2


5 ( x+6 )=2( x+3)

Solution :

5 ( x+6 )=2( x+3)

5 x+30=2 x+6

5 x+30−30=2x+6−30

5 x=2 x−24

5 x=2 x−24

5 x−2 x=2x−2x−24


3 x=−24

3x3

=−243

x=−8

:̌

5 (−8+6 )=2(−8+3)

5(−2)=2(−5)

−10=−10✓

The solution is−8.

Question 3


4 x+3=2(2 x+1)

Solution :

4 x+3=2(2 x+1)

4 x+3=4 x+2

4 x+3−4 x=4 x+2−4 x

3=2

No solution!

Question 4


−3 ( x−2 )+4=5 ( x+2 )−8x

Solution :

−3 x+6+4=5 x+10−8x

−3 x+10=−3 x+10

−3 x+10+3 x=−3 x+10+3 x

10=10

Thereare infinitelymany solutions!


Question 5:

The triangles shown have the same perimeter. What is the value of x?

Solution :

3 ( x+2 )=2 (3x−1 )+ x

3 x+6=6 x−2+ x

3 x+6=7 x−2

3 x+6−6=7 x−2−6

3 x=7 x−8

3 x−7 x=7 x−8−7 x

−4 x=−8

−4 x−4

=−8−4

x=2

:̌

3 (2+2 )=2 (3(2)−1 )+2

3 (4 )=2 (6−1 )+2

12=2 (5 )+2

12=12✓

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Algebra 1 Lesson 1-6 Solving for a variable


Question 1:

Which equationis the result of Solving9+3 x=2 y for x?

A.9+3 y

2=x

B.23y−9=x

C. x=23y−3

D. x=2 y−3

Question 2:

Solve for the indicated variable

a¿ y=mx+b for x

y−b=mx

x= y−bm

b) PV=nRT for T

T=PVnR

c) A=12bh for b

bh=2 A

b=2Ah

Question 3:

The formula for a Fahrenheit temperature in terms of degree – Celsius is


F=95C+32Solve for C

F−32=95C

9C=5 (F−32 )

C=59(F−32)

Question 4:

In your own words, explain how to solve a literal equation for one of the variables

Use inverse operations to isolate the indicated variable on one side of the equation. Be sure the variable is the only expression on one side of the equation and does not appear on the other side

Question 5:

Solve for the indicated variable

v2=u2+2as for s

v2−u2=2as

s= v2−u2

2a

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Algebra 1 Lesson 1-7 Solving absolute value equations

Question 1:

forwhichof the followingis n=−3asolution ?

A. |n−1|=2

B. |n+2|=−1

C. |n−2|=1

D. |n+1|=2

Question 2:

Solve each equation

a¿|2 x−4|=22

2 x−4=±22

2 x−4=22∨2x−4=−22

2 x=26∨2x=−18

x=13∨x=−9

b) |x−3|+14=5

|x−3|=−9 No solution

Question 3:

Is there a value of a for which the equation |x−a|=1 has exactly one solution? Explain

No, no matter what value of a is chosen, there will always be two solutions a+1 and a-1

Question 4:


The perimeter of a rectangle is 100 inches. The length of the rectangle is |2 x−4| inches and the width is x inches. What are the possible values of x?

P=2 L+2W

100=2|2 x−4|+2 x

50=|2x−4|+x

|2 x−4|=50−x

2 x−4=50−x∨2 x−4=−50+x

3 x=54∨x=−46 rejected

x=18

Question 5:

A thermostat is set so that the temperature in a laboratory freezer stays within 2.5 ° F of 2 °F write and solve an absolute value equation to find the maximum and minimum temperatures in the freezer

|x−2|=2.5

x−2=2.5∨x−2=−2.5

x=4.5° F∨x=−0.5 ° F

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Algebra 1 Lesson 1-8 Rates, Ratios and proportions

Question 1:

Ali walksat a speed of 4miles per hour .Hewalks for 20minutes∈a straight line at this rate .approximatly what distancedoes Ali walk ?

A. 0.06miles

B. 1.3miles

C. 5miles

D. 80miles

Question 2:

Solve each proportion

a)x−1

3= x+1

55 x−5=3x+32 x=8x=4

b)3x+5

14= x

39 x+15=14 x5 x=15x=3

c)1x= 1

6 x−16 x−1= x5 x=1

x=15

Question 3:


The ratio of faculty members to students at a college is 1:15 there are 675 students. How many faculty members are there?

Let x be the number of faculty members

115

= x675

15 x=675

x=45

There are 45 faculty members

Question 4:

On a certain day, the exchange rate was 60 U.S. dollars for 50 euro. How many U.S.dollars were 70 euro worth that day.

$60→50euro

$ x→70euro

50 x=60 (70 )

x=60(70)50

=$ 84

Question 5:

Give three Examples of proportions, how do you know they are proportions?

12=3

6

14= 5

20

37= 6

14

They are proportions because the cross products are equal

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Algebra 1 Lesson 1-9 Applications of proportions

Question 1:

Abeachball holds 800 cubic inchesof air . Another beachball hasa radius that is halfthat of thelarger ball . Howmuchair does the smaller ball hold?

A. 400 cubicinches

B. 200 cubic inches

C. 100 cubic inches

D. 80cubic inches

Question2:

A triangle has side lengths of 5 inches, 12 inches and 15 inches. Every dimension is

multiplied by 15 to form a new triangle. How is the ratio of the perimeters related to the

ratio of the corresponding sides?

Perimeter of the original triangle = 5+12+15=32inches

The new triangle has side lengths 1,2.4 3

Perimeter of the new triangle = 15(5+12+15)=6.4 inches

ratios of perimeters= 326.4

ratios of corresponding sides=51= 12

2.4=15

3= 32

6.4

The ratio of the perimeters is equal to the ratio of the corresponding side lengths


Question3:

Ahmedis 5 feet tall∧casts a shadow3.5 feet long . At the same time , the flagpole

outside his school casts a shadow 14 feet long .write∧solve a proportion ¿ find theheight

of the flagpole

Let h be the height of the flagpole

5h=3.5

14

3.5h=70

h= 703.5

=20 feet

Question4:

A rectangle has length 5 cm and width 3 cm. A similar rectangle has length 7.25 cm what is the width of this rectangle

Let x be the width of the rectangle

57.25

=3x

21.75=5x

x=21.755

=4.35

Question5:

At Pizza Palace, a pizza with diameter of 8 inches cost $ 6. The restaurant manager says that a 16 inch pizza should be priced at $ 12 because it is twice as large. Do you agree?

No, a 16 inch pizza actually has 4 times the area so the cost should be 4 times as much

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Algebra 1 Lesson 1-10 Precision and accuracy

Question 1:

Themassof crystal is0.9728gramswhat is themass of the crystal ¿ the nearestmilligram

A. 1milligram

B. 9.73milligram

C. 973milligram

D. 972.8milligram

Question 2:

Round each measurement to the specified precision

a) 5456.3 mi to the nearest mile : 5456 mi b) 3.627 m to the nearest hundredth of meter : 3.63 m c) 119.8 ft to the nearest ten feet : 120 feet d) 62.301 cg to the nearest tenth of a centigram : 62.3 cg

Question 3:

Choose the more precise measurement in each pair

a) 16.270 Liters , 16453.2mlb) 437 cm , 437 mmc) 33mg , 0.033 g neitherd) 67 min , 1.1 h

Question 4:

Rewrite each specified tolerance as a percent

a) 25in ±0.25∈¿ : 25∈±1%b) 240 ft ±12 ft: 240 ft ±5 %

Question 5:

Write the possible range of each measurement. Round to the nearest hundredth

a¿50m±4 %: 48.00−52.00m

b¿90 ° F ±15 % :76.50F−103.50 F


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