ausd ms algebra i first trimester study guide€¦ · ausd ms algebra i first trimester study guide...

AUSD MS Algebra I First Trimester Study Guide

of 9 MCC@WCCUSD (AUSD) 10/29/12

1 Evaluate x y2 − 4( ) ÷ 2x + y( ) when x = −6 and y = −3 . x y2 − 4( ) ÷ 2x + y( )

= −6( ) −3( )2 − 4"#

$% ÷ 2 −6( ) + −3( )"# $%

= −6( ) 9 − 4( ) ÷ 2 −6( ) + −3( )"# $%

= −6( ) 5( ) ÷ −12( ) + −3( )"# $%= −6( ) 5( ) ÷ −15[ ]= −30 ÷ −15( )= 2

1.0/6.EE.2c

2 Simplify −7x + 3− 4x − 8 by combining like

terms. − 7x + 3− 4x − 8

= −7x − 4x + 3− 8= −7x − 4x( ) + 3− 8( )= −11x + −5( )= −11x − 5

1.0, 1.1/6.EE.3

3 Solve 3x + 2 = 12x − 8 − 4x .

3x + 2 = 12x − 8 − 4x3x + 2 = 12x − 4x − 83x + 2 = 8x − 8

3x + 2 − 3x = 8x − 8 − 3x2 = 8x − 3x − 82 = 5x − 8

2 + 8 = 5x − 8 + 810 = 5x105=5x5

2 = x

5.0/8.EE.7

1′ You Try:

Evaluate s2 −152t − 7

when s = −7 and t = 12 .

2′ You Try:

Simplify 3 −2x + 4( ) + 8x − 7 .

3′ You Try:

Solve 4 − 2m = 7 − 3m



4 Solve −3 8p −10( ) = −12p + 6 p − 4( ) .

−3 8p −10( ) = −12p + 6 p − 4( )−24 p + 30 = −12p + 6p − 24−24 p + 30 = −6p − 24

−24 p + 30 + 6p = −6p − 24 + 6p−24 p + 6p + 30 = −6p + 6p − 24

−18p + 30 = −24−18p + 30 − 30 = −24 − 30

−18p = −54−18p−18

=−54−18

p = 3

5.0/8.EE.7

5 Solve 6 2x − 4( ) ≤ 2 4x + 6( ) .

6 2x − 4( ) ≤ 2 4x + 6( )12x − 24 ≤ 8x +12

12x − 24 − 8x ≤ 8x +12 − 8x12x − 8x − 24 ≤ 8x − 8x +12

4x − 24 ≤ 124x − 24 + 24 ≤ 12 + 24

4x ≤ 364x4≤364

x ≤ 9

4.0, 5.0/N.RN.2

4′ You Try: Solve 4 2m + 5( ) = 3m − 5

5′ You Try:

Solve 4(2𝑝 − 2) ≥ 2(𝑝 − 1)



6 Solve 2p − 5p + 8 > −22

2p − 5p + 8 > −22−3p + 8 > −22

−3p + 8 − 8 > −22 − 8−3p > −30−3p−3

<−30−3

p < 10

4.0, 5.0/A.REI.3

7 Evaluate 4x4 when x = −3.

4x4

= 4 −3( )4

= 4 81( )

= 22 i92

= 22 i 92

= 2i9= 18

1.0, 2.0/N.RN.2

6′ You Try: Solve 7x − 3x − 9 < 27 .

7′ You Try:

Evaluate −2x3 when x = −2



8 Find the Domain and Range of the relation: 0,1( ), 1, 3( ) 2,5( ) 3,7( ){ }

The Domain of a relation represented by a set of ordered pairs x, y( ) is the set of all x values. The Range of a relation represented by a set of ordered pairs x, y( ) is the set of all y values. ∴ Domain: 0,1,2,3{ } ; Range: 1,3,5,7{ } Is the above relation a function? A relation is a function if for each domain element there is exactly one range element. ∴ 0,1( ), 1, 3( ), 2,5( ), 3, 7( ){ } is a function.

16.0, 17.0, 18.0/8.F.1

9 Solve the equation for y .

15x + 5y = 20

15x + 5y −15x = 20 −15x5y = −15x + 205y5=−15x5

+205

y = −3x + 4

2.0/A.REI.3

8′ You Try: A) Find the Domain and Range of the relation: 0,−1( ), 1,−4( ), 8,−25( ), 9,−28( ){ }

B) Is the above relation a function?

9′ You Try:

Solve the equation forn : −3m − 6n = −12



10 Find the slope of the line that passes through the points −2,4( ) and 5,−3( ) . Let −2,4( ) be Point 1 Let 5,−3( ) be Point 2

m =y2 − y1x2 − x1

m =−3− 45 − −2( )

m =−77

m = −1

∴ The slope of the line is -1

AF 3.3/8.EE.6

11 Find the x -intercept and the y -intercept of

the line whose equation is −4x + 8y = 16 . To find the x -intercept, let y = 0 and solve for x . −4x + 8y = 16

−4x + 8 0( ) = 16−4x = 16x = −4

∴ the x -intercept is −4,0( ) To find the y -intercept, let x = 0 and solve for y . −4x + 8y = 16

−4 0( ) + 8y = 168y = 16y = 2

∴ the y -intercept is 0,2( )

6.0/F.IF.4

10′ You Try: Find the slope of the line that passes through the points 3,−8( ) and −2,−4( ) .

11′ You Try:

Find the x -intercept and the y -intercept of the line whose equation is 5x − 4y = −20 .



12 Is the ordered pair −3,2( ) a solution of the linear equation 2x + 5y = 4 ? Substitute −3 for x and 2 for y . If these substitutions satisfy the equation (make it true), then the ordered pair is a solution.

2x + 5y = 42 −3( ) + 5 2( ) = 4

−6 +10 = 44 = 4

∴ −3,2( ) is a solution. The point represented by −3,2( ) lies on the graph of the equation 2x + 5y = 4 .

6.0, 7.0/A.REI.10

End of Study Guide

12′ You Try: Is the ordered pair 3,1( ) a solution of the linear equation y = −2x + 7 ?



You Try Solutions:

1´ You try:

Evaluate s2 −152t − 7

when s = −7 and t = 12 .

=−7 ! − 152 12 − 7

=49− 1524− 7

=3417

= 2

2´ You try:

Simplify 3 −2x + 4( ) + 8x − 7 . = −6𝑥 + 12+ 8𝑥 − 7 = −6𝑥 + 8𝑥 + 12− 7 = 2𝑥 + 5

3´ You try:

Solve 4 − 2m = 7 − 3m 4− 2𝑚 + 0 = 4+ 3− 3𝑚 −2𝑚 = 3− 3𝑚 −2𝑚 + 3𝑚 − 3𝑚 = 3− 3𝑚 𝑚 = 3

4´ You try: Solve 4 2m + 5( ) = 3m − 5 8𝑚 + 20 = 3𝑚 − 5 3𝑚 + 5𝑚 + 20 = 3𝑚 − 5+ 20− 20 5𝑚 = −25 !!

!= !!"

!

𝑚 = −5

5´ You try:

Solve 4(2𝑝 − 2) ≥ 2(𝑝 − 1) !(!!!!)

!≥ !(!!!)

!

2(2𝑝 − 2) ≥ 𝑝 − 1 4𝑝 − 4 ≥ 𝑝 − 1 3𝑝 + 𝑝 − 4 ≥ 𝑝 − 1+ 4− 4 3𝑝 ≥ 3 𝑝 + 𝑝 + 𝑝 ≥ 1+ 1+ 1 𝑝 ≥ 1



6´ You try: Solve 7x − 3x − 9 < 27 . 4𝑥 − 9 < 27 4𝑥 − 9+ 9 < 27+ 9 4𝑥 < 36 !!

!< !"

!

𝑥 < 9

7´ You try:

Evaluate −2x3 when x = −2 = −2(−2)! = −2 −2 −2 −2 = 16 = 4

8´ You try:

A) Find the Domain and Range of the relation: 0,−1( ), 1,−4( ), 8,−25( ), 9,−28( ){ }

Domain 0,1,8,9 Range −28,−25,−4,−1 B) Is the above relation a function? Yes it is a function because each input has exactly one output.

9´ You try: Solve the equation forn : −3m − 6n = −12 −3m+ 3m− 6n = −12+ 3m −6n = 3m− 12 !!"

!!= !"!!"

!!

n = !"

!!− !"

!!

n = − !

!𝑚 + 2

10´ You try:

Find the slope of the line that passes through the points 3,−8( ) and −2,−4( ) . Let (−2,−4) be Point 1. Let 3,−8 be Point 2. m = !!!!!

!!!!!

m = !! !(!!)

! !(!!)

m = !!

!



11´ You try: Find the x -intercept and the y -intercept of the line whose equation is 5x − 4y = −20 .

12´ You try:

Is the ordered pair 3,1( ) a solution of the linear equation y = −2x + 7 ? 1 = −2 3 + 7 1 = −6+ 7 1 = 1 ∴ 3,1 is a solution.

Let 𝑦 = 0 5𝑥 − 4(0) = −20 5𝑥 = −20 !!

!= !!"

!

𝑥 = −4 ∴ 𝑥 − intercept is (−4,0)

Let 𝑥 = 0 5(0)− 4𝑦 = −20 −4𝑦 = −20 !!!

!!= !!"

!!

𝑦 = 5 ∴ 𝑦− intercept is (0,5)

ausd ms algebra i first trimester study guide€¦ · ausd ms algebra i first trimester study guide...

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