1.compare (36 + 64)^ 1/2 and (36)^ 1/2 + (64)^ 1/2 (100)^ 1/2 and 6 + 8 10 < 14 2.compare (144 +...

1. Compare

(36 + 64)^1/2 and (36)^ 1/2 + (64)^ 1/2

(100)^ 1/2 and 6 + 8

10 < 14

2. Compare

(144 + 25)^1/2 and (144)^1/2 + (25)^1/2

(169)^1/2 and 12 + 5

13 < 17

3. Compare

(9 + 16)^1/2 and (9)^1/2 + (16)^1/2

(25)^1/2 and 3 + 4

5 < 7

A1.1.1 Compare real number expressions.

A1.1.2 Simplify square roots using factors.

1. Explain why (48)1/2 = 4*(3)1/2

(48)1/2 = (16)1/2 *(3)1/2 = 4*(3)1/2

2. Explain why (125)1/2 = 5*(5)1/2

(125)1/2 = (25)1/2 *(5)1/2 = 5*(5)1/2

3. Explain why (24)1/2 = 2*(6)1/2

(24)1/2 = (4)1/2 *(6)1/2 = 2*(6)1/2

A1.1.3 Understand and use the distributive,associative, and commutative properties.

1. (6x2 - 5x - 1) - 2(x2 - 3x - 4) = 6x2 - 5x - 1 - 2x2 + 6x + 8 Distributive

= 6x2 - 2x2 – 5x + 6x – 1+ 8 Associative & Commutative

= 4x2 + x + 72. (6x2 - 5x - 1) + 2(x2 - 3x - 4) = 6x2 - 5x - 1 + 2x2 - 6x – 8 Distributive

= 6x2 + 2x2 - 5x - 6x - 1 – 8 Associative & Commutative

= 8x2 - 11x – 93. 3(4x2 - 3x + 1) - (x2 +2x - 6) = 12x2 - 9x + 3 - x2 - 2x – 6 Distributive

= 12x2 - x2 - 9x - 2x + 3 – 6 Associative & Commutative

= 11x2 - 11x – 3

A1.1.4 Use the laws of exponents for rational exponents.

Simplify.

1. 253/2 = [(25)½]3 = (5)3 = 125

2. 363/2 = [(36)½]3 = (6)3 = 216

3. 36(-1/2) = [(36)½](-1) = (6)-1 = 16

A1.1.5 Use dimensional (unit) analysis to organize conversions and computations.

1. Convert 5 miles per hour to feet per second.

2. Convert a 12 ft. by 9 ft. by 3 in. deep to cubic yards.

Find the volume by converting all measurements to yards. 12 ft = 4 yards, 9 feet = 3 yards, 3 in = ¼ feet = 1/12

yards Multiply 4 yards. * 3 yards * (1/12) yards = 1 cubic yard3. Convert 30 cents per minute to dollars per hour. 30 cents * 1 dollar * 60 minutes = $18/hour 1 minute 100 cents 1 hour

5 miles * 5280 feet * 1 hour * 1 minute = 43 feet = 7.17 ft/sec.1 hour 1 mile 60 minutes 60 seconds 6 sec.

miles to ft hours to min. min. to sec. reduce approximate

A1.2.1 Solve linear equations.

Solve the equation:

1. 7a + 2 = 5a - 3a + 8

7a + 2 = 2a + 8

-2a -2a

5a + 2 = 8 -2 -2

5a = 6 5 5

a = 6/5 or 1.2

2. 7x - 2 + 3x = 4x - 3 + 8

10x – 2 = 4x + 5

-4x -4x

6x – 2 = 5 + 2 +2

6x = 7 6 6

x = 7/6, 1 1/6, or 1.167

3. 5y - 2 - y = 4y - 3 + 1

4y – 2 = 4y – 2 -4y -4y

- 2 = -2

This is an identity, all real numbers are solutions

A1.2.2 Solve equations and formulas for a specified variable.

1. q = 4p - 11 for p +11 +11

q + 11 = 4p 4 4

q/4 + 11/4 = p

2. A = 2l + 2w for w. -2l -2l

A – 2l = 2w 2 2

A/2 – 2l/2 = w

A/2 – l = w

3. 5(2x + 4) = 3y – 2x for x.

10x + 8 = 3y – 2x +2x +2x

12x + 8 = 3y -8 -8

12x = 3y – 8 12 12

x = 3y/12 – 8/12

x = y/4 – 2/3

A1.2.3 Find solution sets of linear inequalities when possible numbers are given for the variable.

1. Solve the inequality 6x - 3 > 10, for x in the set {0, 1, 2, 3, 4}. for x = 0, 6*0 – 3 > 10, so 0 – 3 > 10, implies -3 > 10, not true. Not a solution. for x = 1, 6*1 – 3 > 10, so 6 – 3 > 10, implies 3 > 10, not true. Not a solution. for x = 2, 6*2 – 3 > 10, so 12 – 3 > 10, implies 9 > 10, not true. Not a solution. for x = 3, 6*3 – 3 > 10, so 18 – 3 > 10, implies 15 > 10, true. x = 3 is a solution. for x = 4, 6*4 – 3 > 10, so 24 – 3 > 10, implies 21 > 10, true. x = 4 is a solution.

2. Solve the inequality 15 > 3x - 12, for x in the set {7, 9, 11, 13}. for x = 7, 15 > 3*7 – 12, so 15 > 21 – 12, implies 15 > 9, true. x = 7 is a solution. for x = 9, 15 > 3*9 – 12, so 15 > 27 – 12, implies 15 > 15, not true. x = 9 is not a solution. for x = 11, 15 > 3*11 – 12, so 15 > 33 – 12, implies 15 > 21, not true. x = 11 is not a solution. for x = 13, 15 > 3*13 – 12, so 15 > 39 – 12, implies 15 > 27, not true. x = 13 is not a solution.

3. Solve the inequality 2 – 5x < 12, for x in the set {-1, -2, -3, -4}. for x = -1, 2 – 5(-1) < 12, so 2 + 5 < 12, implies 7 < 12, true. x = -1 is a solution. for x = -2, 2 – 5(-2) < 12, so 2 + 10 < 12, implies 12 < 12, true. x = -2 is a solution. for x = -3, 2 – 5(-3) < 12, so 2 + 15 < 12, implies 17 < 12, not true. x = -3 is not a solution. for x = -4, 2 – 5(-4) < 12, so 2 + 20 < 12, implies 22 < 12, not true. x = -4 is not a solution.

A1.2.4 Solve linear inequalities using properties of order.

1. Solve the inequality 8x - 7 < 2x + 5, explaining each step in your solution. 8x – 7 < 2x + 5, original problem -2x -2x , remove 2x from each side 6x – 7 < 5 simplify each side +7 +7 add 7 to each side to get the x-term (6x) by itself. 6x < 12 simplify each side. 6 6 divide both sides by the coefficient of the variable term 2 > x or x < 2 simplify each side

2. Solve the inequality 3y + 2 > 5y + 10, explaining each step in your solution 3y + 2 > 5y + 10, original problem -3y -3y , remove 3y from each side 2 > 2y + 10, simplify each side -10 -10 , remove 10 from the side with the variable -8 > 2y simplify each side 2 2 divide both sides by the coefficient of the variable y < -4 or -4 > y simplify each side

3. Solve the inequality 3a - 14 5a + 10, explaining each step in your solution. 3a - 14 5a + 10, original problem -3a -3a , remove 3a from each side to remove all a’s from one side -14 2a + 10 , simplify each side -10 -10, remove 10 from each side to get the variable term by itself -24 2a simplify each side 2 2 divide both sides by the coefficient of the variable term a -12 or -12 a , simplify each side

A1.2.5 Solve combined linear inequalities.

1. Solve the inequalities -7 < 3x - 5 < 11. -7 < 3x – 5 and 3x – 5 < 11 +5 +5 +5 +5 -2 < 3x and 3x < 16 3 3 3 3 -2/3 < x and x < 16/3

-2/3 < x < 16/3

2. Solve the inequalities 10 < -3x - 5 < 28.

10 < -3x – 5 and –3x – 5 < 28

+5 +5 +5 +5

15 < -3x and -3x < 33

-3 -3 -3 -3

Remember to flip the inequality when multiplying or dividing by a negative number.

so, -5 > x and x > -11

-5 > x > -11

3. Solve the inequalities -4 < 2x + 6 < 4.

-4 < 2x + 6 and 2x + 6 < 4 -6 -6 -6 -6 -10 < 2x and 2x < -2 2 2 2 2

-5 < x and x < -1

-5 < x < -1

A1.2.6 Solve word problems that involve linear equations, formulas, and inequalities.

1. You are selling tickets for a play that cost $3 each. You want to sell at least $50 worth. Write and solve an inequality for the number of tickets you must sell.

3 times the number of tickets is greater than 50 or mathematically ---- 3n > 50

Solve 3n > 50 3 3

n > 16.67, which means you need to sell at least 17 tickets or more than 16.

2. You are filling a 20 gallon bucket with water at 1.5 gallons per minute. Write and solve an inequality for the number of minutes needed to have a bucket full.

1.5 gallons times the number of minutes must equal 20 gallons, or mathematically ---- 1.5*n = 20

Solve 1.5*n = 20 1.5 1.5

n = 13.3333………, round up to 14 minutes since the bucket will not be full at 13 minutes.3. You are needing to gross a minimum of $214 this week, your hourly rate is $4.75. Write and solve an inequality for the number of hours you must work.

4.75 times the hours worked must equal 214, or mathematically ---- 4.75*n = 214

Solve 4.75*n = 214 4.75 4.75

n = 45.05, round up to 46 hours or ask your boss to pay you for 3 minutes of work.

A1.3.1 Sketch a reasonable graph for a given relationship.

1. Sketch a reasonable graph for a person’s height from age 0 to 25.

2. Sketch a reasonable graph for the continuous speed of a school bus’s trip to school.

3. Sketch a reasonable graph for the continuous height off the ground of a person jumping rope.

height in 1 foot increments

Age in 5 year increments

Miles per hour are the vertical axis and time is the horizontal axis. The mph return and stay at zero every time the bus stops for a student.

Height in inches is the vertical axis, time is the horizontal axis. The jumper returns and stays on the ground for a brief moment before jumping again.

A1.3.2 Interpret a graph representing a given situation.

Jessica is riding a bicycle. The graphs below show her speed as it relates to the time she has spent riding. Describe what might have happened to account for the specified graph.

1. Black

2. Green

3. Red

speed

time

1. It appears as if Jessica starts out going downhill since she gains speed rapidly. Then has a few small hills to go over until reaching a bigger hill, after which she levels off.

2. Jessica appears to start on flat ground or even a small downhill grade. The she has a long stretch in which she must be going a little uphill since she gradually loses speed. Then she crests the hill and starts on the downhill to finish her trek.

3. Definitely appears to start on a slightly uphill or flat grade before reaching a big long downhill ride. At the bottom she has to go up a short, but very steep hill. Then she appears to be going downhill again.

A1.3.3 Understand the concept of a function, decide if a given relation is a function, and link equations to functions.

Use paper to generate a list of values for x and y in the following equation.

Based on your data, make a conjecture about whether or not this relation is a function. Explain your reasoning.

1. y = x2 x = y2 y = 3x3 – 4

Pick the x-value (5 different) pick the y-value (5 different) pick the x-value

Try 2 negatives, zero and 2 positive numbers) (Try 2 negatives, zero and 2 positive numbers) (Try 2 negatives, zero and 2 positive numbers)

x y x y x y

-1 1 1 -1 -1 -7

-2 4 4 -2 -2 -28

0 0 0 0 0 0

1 1 1 1 1 -1

2 4 4 2 2 20

Function, each x-value equals a Not a Function, when the x-value equals Function, each x-value equals different y – value. 1 the y-value equals 1 or –1. a different y-value.

A1.3.4 Find the domain and range of a relation.

Find the domain and range of the following relations, explain how can you decide whether you are correct:

1. y = x2

Let x be zero, positive, then negative. Repeat the process for y. Check to insure that all values are possible. Use deductive reasoning to determine which values do not appear to be possible. Correctness could be determined by using a graphing calculator.

2. x = y2


3. 3x = 4y


x y x y

0 0 0 0

1 1 -1,1 1

-1 1 no -1

x y x y

0 0 0 0

1 -1,1 1 1

-1 no 1 -1

x y x y

0 0 0 0

1 ¾ 4/3 1

-1 -¾ -4/3 -1

domain (x-values) can be all real #’s

range (y-values) are all non-negative #’s

domain (x-values) can be all non- negative numbers

range (y-values) can be all real #’s

domain (x-values) can be all real #’s

range (y-values) can be all real #’s

A1.4.1 Graph a linear equation.

1. Draw the graph of the line with slope 3 and y-intercept -2.

2. Draw the graph of the line that passes through the points (3,2) and (–3,-2)

3. Draw the graph of the line with no slope and passes through the point (4,0)

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y-axis

x-axis

y-axis

x-axis

y-axis

A1.4.2 Find the slope, x-intercept and y-intercept of a line given its graph, its equation, or two points on the line.1. Find the slope and y-intercept of the line 4x - 6y =

12. Easiest solution is to put in slope-intercept

form (y=mx + b)

2. Find the slope and y-intercept of the line that passes through (2,-1) and (-2,1).

Use the formula for slope, m = (y2 – y1)/(x2 – x1)

m = [1 – (-1)]/(-2 – 2) or 2/-4 = - ½ Slope is – ½

To find the y-intercept (b), plug the newly found slope and 1 point into the slope-intercept formula of y = mx + b

Use the first point (2, -1) the slope of – ½ and find b.

-1 = - ½*(2) + b -1 = -1 + b +1 +1 0 = b, the y-intercept is zero.

3. Find the slope and y-intercept of the graph.

By looking at the graph the line crosses the y-axis at (0,1) so the y-intercept is 1. Find the slope using rise over run. The rise is 1 and the run is –2. The slope is – ½.

4x – 6y = 12-4x -4x -6y = -4x – 12 -6 -6

y = 2/3x + 2

m = 2/3 and the y-intercept is 2

A1.4.3 Write the equation of a line in slope-intercept form. Understand how the slope and y-intercept of the graph are related to the equation.

1. Write the equation of the line 4x + 6y = 12 in slope-intercept form. What is the slope of this line? Explain your answer.

4x + 6y = 12 -4x -4x

6y = -4x + 12 6 6

y = -2/3x + 2

The slope is -2/3

2. Write the equation of the line2x - y = 4 in slope-intercept form. What is the slope and y-intercept of this line? Explain your answer.

2x – y = 4 -2x -2x - y = -2x + 4

-1 -1

y = 2x – 4

the slope is 2 and the y-intercept is –4.

3. Write the equation of the line 2x – 3y = -5 in slope-intercept form. What is the y-intercept of this line? Explain your answer.

2x – 3y = -5 -2x -2x -3y = -2x – 5

-3 -3

y = 2/3x + 5/3

the y-intercept is 5/3

A1.4.4 Write the equation of a line given appropriate information.

1. Find an equation of the line through the points (1,4) and (3,10), then find an equation of the line through the point (1, 4) perpendicular to the first line.

3. Find an equation of the line through the points (1,2) and (-4,-3), then find an equation of the line through (0,1) perpendicular to the first line.

2. Find an equation of the line through the points (2,3) and (3,2), then find an equation of the line through the origin parallel to the first line.

Find the slope of the first line:m=(y2 – y1)/(x2 – x1)m= (10 – 4) / (3 – 1)m = 6 / 2 = 3

Use slope and the first point to find the y-intercept using y = mx + b

4 = 3*(1) + b4 = 3 + b-3 -31 = b, the y-intercept is 1. Write the equation of the line in slope-intercept form.y = 3x + 1. The slope of this line is 3. All lines perpendicular to this line have a slope that is the negative reciprocal of 3, which is –1/3. Use this slope and the point of (1,4) to find the y-intercept of the perpendicular line. y = mx + b 4 = (–1/3)(1) + b +1/3 +1/3

13/3 = bUse the slope of -1/3 and y-int of 13/3 in the slope-intercept formula to write the equation of the perpendicular line: y = -1/3x + 13/3

Find the slope of the first line:m=(y2 – y1)/(x2 – x1)m= (2 – 3) / (3 – 2)m = -1 / 1 = -1


3 = -1*(2) + b3 = -2 + b+2 +25 = b, the y-intercept is 1. Write the equation of the line in slope-intercept form.y = -x + 5. The slope of this line is -1. All lines parallel to this line have the same slope as this line, which is -1. Use this slope and the point of (0,0) to find the y-intercept of the parallel line. Since this point is on the y-axis the y-intercept is 0.

Use the slope of -1 and y-int of 0 in the slope-intercept formula to write the equation of the parallel line: y = -x + 0 or y = -x

Find the slope of the first line:m=(y2 – y1)/(x2 – x1)m= (-3 – 2) / (-4 – 1)m = -5 / -5 = 1


2 = 1*(1) + b2 = 1 + b-1 -11 = b, the y-intercept is 1. Write the equation of the line in slope-intercept form.y = x + 1. The slope of this line is 1. All lines perpendicular to this line have a slope that is the negative reciprocal of 1, which is –1. Use this slope and the point of (0,1) to find the y-intercept of the perpendicular line. (0,1) is on the y-axis, therefore the y- intercept is 1Use the slope of -13 and y-int of 1 in the slope-intercept formula to write the equation of the perpendicular line: y = -x + 1.

A1.4.5 Write the equation of a line that models a data set and use the equation (or the graph of the equation) to make predictions. Describe the slope of the line in terms of the data, recognizing that

the slope is the rate of change.

1. As your family is traveling along an interstate, you note the distance traveled every 5 minutes. A graph of time and distance shows that the relation is approximately linear. You find that the car travels 2.5 miles every 5 minutes. Write the equation of the line that fits your data. Predict the time for a journey of 50 miles. What does the slope represent?

Let y be the distance traveled.

Let x be the number of 5 minute intervals.

2.5 miles/interval will be the rate traveled .

D = r * t

the equation is y = 2.5*x

To predict a journey (distance) of 50 miles, plug 50 in for y. Solving for x will give you the number of 5 minute intervals for this journey.

50 = 2.5* x 2.5 2.5

20 = xSince 20 is the number of 5-minute intervals, the trip will actually take 100 minutes.

A1.4.6 Graph a linear inequality in two variables.

Draw the graph of the inequality 6x + 8y < 24. Solve for y

6x + 8y < 24-6x -6x

8y < -6x + 24 8 8

y < - ¾ x + 3, the origin is on the graph.

2. Draw the graph of the inequality 2x – 3y > 6

solve for y2x – 3y > 6-2x -2x -3y > -2x + 6 -3 -3 flip

inequality y < 2/3 x – 2, the origin is not on

the graph

3. Draw the graph of the inequality 6x + 2y < -5 solve for y 6x + 2y < -5 -6x -6x

2y < -6x + 5 2 2 y < -3x + 5/2, the origin is on the graph

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x-axis

y-axis

x-axis

y-axis

A1.5.1 Use a graph to estimate the solution of a pair of linear equations in two variables.

1. Graph the equations 3y - x = 0 and 2x + 4y = 16 to find where the lines intersect.

solve each equation for y. 3y – x = 0 and 2x + 6y = 24 +x +x -2x -2x

3y = x 6y = -2x + 24 3 3 6 6 y = 1/3 x y = - 1/3 x + 16 m = 1/3 b = 0 m = - 1/3 b= 4

3. Graph the equations 2x + 4y = 6 and x + y = -3 to find where the lines intersect.

Solve each equation for y.2x + 4y = 6 and x +

y = -1 -2x -2x -x

-x 4y = -2x + 6

y = -x - 1 4 4

y = - ½ x + 3/2

m=- ½ b = 3/2 m=-1 b= -1

2. Graph the equations y = 2x – 5 and y = -½x + 2 to find where the lines

intersect.Each equation is already solved for y.

y = 2x – 3 and y = -½x + 2

m=2 b=-3 m=- ½ b=2

x-axis

y-axis

x-axis

y-axis

x-axis

y-axis

Lines intersect at the point (2,6)

Lines intersect at the point (2,1)

Lines intersect at the point (-5,4)

A1.5.2 Use a graph to find the solution set of a pair of linear inequalities in two variables.

1. Graph the inequalities y < 4 and x + y < 5. Shade the region where both inequalities are true.

Set all equations in y=mx + b form

y < 4 and x + y < 5

-x -x y < -x +

5m=0 and b=4 m= -1 and

b =5the origin is on the graph the origin is on the

graph

3. Graph the inequalities 2x–y < 3 and x + y < 6. Shade the region where both inequalities are true.

Set both inequalities in slope-intercept form.

2x – y < 3 x + y < 6

-2x -2x -x -x

-y < -2x + 3 y < -x + 6

-1 -1 y < 2x – 3 y <

-x + 6 m = 2 b = -3 m

= -1 b = 6 The origin is not on the graph The origin is

on the graph

2. Graph the inequalities x > 3 and y > -2. Shade the region where both inequalities are true.

Graph both lines, the first has no slope and the second has a slope of zero.

The origin is not on x > 3, The origin is on the graph of y >-2

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y-axis

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y-axis

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y-axis

A1.5.3 Understand and use the substitution method to solve a pair of linear equations in two variables.

1. Solve the equations y = 2x and 2x + 3y = 12 by substitution.Since the first equation says y = 2x, insert 2x in the place of y in the second equation, then solve for x.

answer: 2x + 3y = 12 We will insert 2x in for y, removing y. 2x + 3(2x) = 12 2x + 6x = 12 y = 2x or 2x + 3y = 12 8x = 12 y = 2(3/2) or 2(3/2) + 3y =

12 8 8 y = 3 3 + 3y =

12 -3 -3 3y = 9 3 3

x = 3/2 or 1 ½ Now plug 3/2 in for x into either equation, either method finds y = 3, the answer is x = 3/2 and y = 3

then solve for y.

2. Solve the equations x = 4 and 3x – 4y = 4 by substitution.Since the first equation says x = 4, insert 4 in the place of x in the second equation, then solve for y.

3x – 4y = 4 We will insert 4 in for x and remove x. 3(4) – 4y = 4

12 – 4y = 4 -12 -12 -4y = -8

-4 -4 y = -2, the answer is x = 4 (we were given this) and y = -2

3. Solve the equations y – 2x = 3 and y – 7 = 4xSince we do not know either x or y, we will have to solve one equation for y before substituting.

y – 2x = 3 Solve for y in the first equation. +2x +2x y = 2x + 3, Now we can substitute 2x + 3 for y into the other (2nd) equation.

y – 7 = 4x, Plug 2x + 3 in for y in the 2nd equation y – 7 = 4x replace x with -2

2x + 3 – 7 = 4x Now solve for x y – 7 = 4(-2) 2x – 4 = 4x y – 7 = -8

-2x -2x +7 +7 -4 = 2x y = -1 2 2

-2 = x, We will now insert –2 into the 2nd equation. The answer is x = -2 and y = -1

A1.5.4 Understand and use the addition or subtraction method to solve a pair of linear equations in two variables.

1. Use subtraction to solve the equations: 3x + 4y = 11, 3x + 2y = 7.

Align equations vertically Since the x terms are alike 3x + 4y = 11 3x + 4y = 11,

put 2 in place of y Subtract the 2nd from the 1st -(3x + 2y = 7) 3x + 4(2) = 11 Now solve for y 2y = 4 3x + 8 = 11

2 2 -8 -8 y = 2, 3x = 3

plug 2 for y into the 1st equation and solve for x 3 3

the answer is x = 1 and y = 2

2. Use addition to solve the equations: 5x – 4y = 2 and 3x + 4y = 145x – 4y = 2, replace x with 2

Align equations vertically 5(2) – 4y = 2Since the y terms are opposites 5x – 4y = 2 10 – 4y = 2

Add the 2nd line with the 1st 3x + 4y = 14 -10 -10Now solve for x 8x = 16 -4y = -8

8 8 -4 -4plug 2 for x into the 1st equation and solve for y x = 2 y = 2

the answer is x = 2 and y = 2

3. Use subtraction to solve the equations: x – 3y = 7 and 2x – 6y = 14Multiply the 1st equation by 2 2(x - 3y = 7)Makes x terms alike, now alignSubtract the 2nd from the first 2x – 6y = 14

Add the 2nd line with the 1st -(2x – 6y = 14)Now solve for x 0 = 0

Since the left side equals the right side this is an IDENTITYThese equations are the same and their graphs are identical.

A1.5.5 Understand and use multiplication with the addition or subtraction method to solve a pair of linear equations in two variables.

1. Use multiplication with the subtraction method to solve the equations: x + 4y = 16, 3x + 2y = -2.

multiply the first equation by 3 so the x-terms are the same 3x + 12y = 48 x + 4y = 16, y = 5Align the equations so the terms match up, then subtract -(3x + 2y = -2) x + 4(5) = 16

10y = 50 x + 20 = 16 10 10 -20 -20

Plug this answer into the 1st equation y = 5 x = -4the solution is x = -4 and y = 5

Use multiplication with the addition method to solve the equations: 3x - y = 1, -2x + 2y = -6.

multiply the 1st equation by 2 so the y-terms are opposites 6x – 2y = 2 3x – y = 1, x = -1Align the equations so the terms match up, then add -2x + 2y = -6 3(-1) – y = 1

4x = -4 -3 – y = 1 4 4 +3 +3

Plug this answer into the 1st equation x = -1 -y = 4 -1 -1 y = -4

the solution is x = -1 and y = -4

3. Use multiplication with the add./subt. method to solve the equations: 5x – 2y = -7, 4x + 3y = -1

multiply the 1st equation by 3 and the 2nd equation by 2. This will make the y-terms opposites. Stack the equations and add.

1st equation times 3 15x – 6y = -21 5x – 2y = -7, x = -12nd equation times 2 8x + 6y = -2 5(-1) – 2y = -7Add the equations vertically 23x = -23 -5 - 2y = -7

23 23 +5 +5Plug the answer into the 1st equation x = -1

-2y = -2 -2 -2 y = -1

the solution is x = -1 and y = -1

A1.5.6 Use pairs of linear equations to solve word problems.

1. The income a company makes from a product can be represented by the equation y = 10.5x and the expenses for that product can be represented by the equation y = 5.25x + 10,000, where x is the amount of the product sold and y is the number of dollars.

How much of the product must be sold for the company to reach the break-even point?

First, for the company to break even, the income (# of dollars received from products sold) must be equal to the expenses of the company. Both equations are linear, therefore the point of intersection is the same as the break-even coach. We can solve either by graphing (numbers are too large to do this easily, but it can be done) or you can solve by using the substitution method. Substitution is much easier and will be the method of choice.

Since y = 10.5*x

we will replace y with 10.5*x in the other equation - y = 5.25x + 10,000

10.5x = 5.25x + 10,000 Solve for x

10.5x = 5.25x + 10,000 - 5.25x -5.25x

5.25x = 10,000 5.25 5.25

x = 1904.76Plug this answer back into the equation above to find y

y = 10.5x replace x with 1904.76

y = 10.5*1904.76 = 20,000

A1.6.1 Add and subtract polynomials.

1. Simplify (4x2 - 7x + 2) - (x2 + 4x - 5).

4x2 - 7x + 2 - x2 - 4x + 5 Remove parentheses, change all signs inside the parentheses if a minus sign is in front.

4x2 - x2 - 7x - 4x + 2 + 5 Use Commutative and Associative Properties to put like terms together (watch signs)

3x2 – 11x + 7 Combine like terms and you are finished.

2. Simplify (5x2 + 3x + 4) + (3x2 - 3x - 6).

5x2 + 3x + 4 + 3x2 - 3x - 6 Remove parentheses, change all signs inside the parentheses if a minus sign is in front.

5x2 + 3x2 + 3x - 3x + 4 - 6 Use Commutative and Associative Properties to put like terms together (watch signs)

8x2 – 2 Combine like terms and you are finished..

3. Simplify (5 – 6x - 5x2) - (8 – 3x - 8x2).

5 – 6x - 5x2 - 8 + 3x + 8x2 Remove parentheses, change all signs inside the parentheses if a minus sign is in front.

- 5x2 + 8x2 – 6x + 3x + 5 - 8 Use Commutative and Associative Properties to put like terms together (watch signs)

3x2 – 6x - 3 Combine like terms and you are finished..

A1.6.2 Multiply and divide monomials.

1. Simplify a2b5/ab2.

2. Simplify (2a2b4)(-3a3b4c)

3.Simplify (-6x2y3z5)/(-15xy4z2)

a2b5 Cancel one “a” and two “b”’s ab2 This will leave one a on top and 3 b’s on top

Nothing will be left on bottom, but because canceling is

division, we will be left with one. When we divide by 1, we do not include the

denominator.

a*b3 This is the answer.

(2a2b4)(-3a3b4c) Multiply to get your solution. When multiplying

monomials, multiply the coefficients (number factors)

and count number of variable factors by adding the

exponents of each factor.

-6a5b8c This is the answer

-6x2y3z5 Rewrite fraction as a stack. Cancel common factors -15xy4z2 of the coefficients (-3) and reduce variable factors by

subtracting the smallest exponent from the larger and

leaving the variable in the location of the larger

exponent.

2xz3 This is the answer 5y

A1.6.3 Find powers and roots of monomials (only when the answer has an integer exponent).

1. Find the square root of a2b6.(a2b6)½ rewrite with the square rootab3

2. Find the cube of 2x2y3 (2x2y3)3 Rewrite by cubing the value (3rd power) (2x2y3) (2x2y3) (2x2y3) Rewrite without exponents8x6y9 Multiply number factors, add exponents

for variable

3. Find the cube root of –125a6b9

(–125a6b9)1/3 Rewrite using the cube root (power of 1/3)

-5a2 b3

A1.6.4 Multiply polynomials.

1. Multiply (n + 2)(4n - 5).

Use FOIL to multiply

First*First + Outside*Outside + Inside*Inside + last*last n*4n + n*(-5) + 2*4n + 2 * (-5) 4n2 + (-5n) + 8n + (-10) 4n2 + (-5n) + 8n + (-10) Rewritten in shorten form

4n2 + 3n –10 Answer in simplified form

2. Multiply (3x2 – 2x + 3)(4x + 1)

Must multiply each term in the first polynomial times each term in the 2nd binomial

3x2 *4x + 3x2 *1 + (-2x)*4x + (-2x)*1 + 3*4x + 3*1 12x3 + 3x2 + (-8x2) + (-2x) + 12x + 3 Simplify

multiplication 12x3 – 5x2 + 10x + 3 Answer after adding

like terms

3. Multiply (8x – 3)(8x + 3)Use FOIL to multiply

First*First + Outside*Outside + Inside*Inside + last*last 8x*8x + 8x*3 + (-3)*8x + (-3)(3) 64x2 + 24x + (-24x) + (-9) 64x2 + 24x + (-24x) + (-9) Rewritten in shorten form

64x2 – 9 Answer in simplified form

A1.6.5 Divide polynomials by monomials.

1. Divide 4x3y2 + 8xy4 - 6x2y5 by 2xy2. Set up as a long division problem

2x2 + 4y2 - 3 xy3 This is the answer

2xy2 4x3y2 + 8xy4 - 6x2y5 - 4x3y2 Subtract and bring down the next term

0 + 8xy4 What times 2xy2 = 8xy4, 4y2

- 8xy4 Subtract and bring down the next term 0 - 6x2y5 What times 2xy2 = 6x2y5, -3xy3

+ 6x2y5 Subtract and bring down the next term 0 No remainder

2. Divide 9a5b3 – 12a4b2 + 3a3b5 by 3a2b2 Set up as a long division problem.

3a7b - 4a2 + ab3 This is your answer

3a2b2 9a5b3 – 12a4b2 + 3a3b5 What times 3a2b2 = 9a9b3, 3a7b - 9a5b3 Subtract and bring down the next

term 0 - 12a4b2 What times 3a2b2 = -12a4b2, -4a2 - 12a4b2 Subtract and bring down the next term

0 + 3a3b5 What times 3a2b2 = 3a3b5 , ab3

- 3a3b5 remainder is zero

3. Divide 8a3b2 + 6a2b3+ 4ab4 by 2ab Set up as a long division problem.

3a7b - 4a2 + ab3 This is your answer

2ab 8a3b2 + 6a2b3+ 4ab4 What times 2ab = 8a3b2, 4a2b - 8a3b2 Subtract and bring down the next

term 0 + 6a2b3 What times 2ab = 6a2b3, 3ab2 - 6a2b3 Subtract and bring down the next term

0 + 4ab4 What times 2ab = 4ab4 , 2b3

- 4ab4 Remainder is zero

A1.6.6 Find a common monomial factor in a polynomial.

1. Factor: 36xy2 + 18xy4 - 12x2y4. Find the GCF of the coefficients. Coefficients are 36, 18, and 12.

2, 3, and 6 are factors of all three, 6 is the greatest

Next look at the variable factors x and yThere is one x factor in the 1st and 2nd terms and two in the 3rd term. “x” is the common factorThere is two y factors in the 1st term and four y factors in the 2nd and 3rd terms, is the y2 common factor

The common monomial factor is 6xy2

2. Factor: 12x3 + 9x2 + 3x Find the GCF of the coefficients. Coefficients are 12, 9, and 3.

3 is a factor of all three, 3 is the greatestNext look at the variable factor x

There are 3 x-factors in the 1st term, 2 in the 2nd, and one in the 3rd, “x” is the common factor

The common monomial factor is 3x

3. Factor: -8x4y2 – 12x3y3 + 20x2y4 Find the GCF of the coefficients. Coefficients are -8, -12, and 20.

2 and 4 are common factors, 4 is the greatest (use –4 to pull out the “-”)

Next look at the variable factors x and yThere are 4 x factors in the 1stterm, 3 in the 2nd, and 2 in the 3rd. “x2” is the common factorThere is two y factors in the 1st term, 3 in the 2nd,and 4 in the 3rd, y2 common factor

The common monomial factor is -4x2y2 or 4x2y2

A1.6.7 Factor the difference of two squares and other quadratics.

1. Factor: 4x2 – 25Reverse FOIL, Find F and L first, then make sure O+I = the middle term (zero in this case)(2x + 5)(2x – 5) are the factors

2. Factor: 6x2 + 5x – 56Reverse FOIL, Find F and L first, then make sure O+I = 5xChoices for F are 6x*1x or 3x*2x, try 3x and 2x 1st

(3x )(2x ), now let’s look at choices for L, 8*7, 4*14, 2*28, and 1*56 (must also reverse)

Try 8,7 first (one of the factors must be negative)(3x + 8)(2x - 7) O+I = -21x + 16x = -5x, this is not right, we want +5x(3x – 8)(2x + 7) change the signs, and O+I = 21x – 16x = 5x, these are the factors

3. Factor: 2x2 - 7x + 3.Reverse FOIL, Find F and L first, then make sure O+I = 5x

Choices for F are 2x*1x(2x )(x ), now let’s look at choices for L, 3*1 (must also reverse)

Try 3*1 first (the factors must be the same)(2x + 3)(x + 1) O+I = 2x + 3x = 5x, this is not right, we want -7x(2x - 3)(x - 1) change the signs, and O+I = -2x – 3x = -5x, not right – try reversing the order (2x – 1)(x – 3) O + I = -6x – 1x = -7x, this is correct and these are the factors

A1.6.8 Understand & describe the relationships among the solutions of an equation, the x-intercepts of a graph, the zeros of a function, and the factors of a polynomial expression.

1. A graphing calculator can be used to solve 3x2 - 5x +1 = 0 to the nearest tenth.

Justify using the x-intercepts of y = 3x2 - 5x + 1 as the solutions of the equation.

The intercepts appear to be at the points (0,0) and (1,0), plug each point back into the original equation to check your answer.

(0,0) 0 = 3*0 – 5*0 + 1, 0 = 1 – this is not a solution

(1,0) 0 = 3*1 – 5*1 + 1, 0 = -1, this is not a solution

A1.7.1 Simplify algebraic ratios.

x2 - 16x2 + 4x1.

3x2 – 4x - 7

x2 - 12.

4x3 + 6x2 – 4x36x2 – 6x - 63.

(x + 4)(x – 4 = x – 4 This is the answer x(x + 4) x

(3x – 7)(x + 1 = 3x – 7 This is the answer (x+1)(x-1) x – 1

2x(2x2 + 3x – 2) = 2x(2x –1)(x + 2) = x(x + 2)6 (6x2 – x – 1) 6 (3x + 2)(2x – 1) 3(3x + 2)

A1.7.2 Solve algebraic proportions.

x + 5 41. = 3x + 5

7

2x + 2 -3

2. =x - 2

-2

3.5x - 3 6

=5x - 5

10

To solve these, cross multiply and then solve for x.

Cross multiply: 7(x + 5) = 4(3x + 5) 7x + 35 = 12x + 20

-7x -7x 35 = 5x + 20 -20 -20 15 = 5x, x = 3

Cross multiply: -2(2x + 2) = -3(x – 2) -4x – 4 = -3x + 6 Watch your signs

+4x +4x -4 = x + 6 -6 -6

-10 = x

Cross multiply: 10(5x – 3) = 6(5x – 5)50x – 30 = 30x – 30-30x -30x20x - 30 = -30 + 30 + 30 20x = 0 x = 0

1.compare (36 + 64)^ 1/2 and (36)^ 1/2 + (64)^ 1/2 (100)^ 1/2 and 6 + 8 10 < 14 2.compare (144 +...

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