A1.1.1 Compare real number expression
1.
Compare
(36 + 64)^1/2 and (36)^ 1/2 + (64)^ 1/2
(100)^ 1/2
and
6 + 8
10 <
2.
14
Compare
(144 + 25)^1/2 and (144)^1/2 + (25)^1/2
(169)^1/2 and
13
<
12 + 5
17
3. Compare
(9 + 16)^1/2 and (9)^1/2 + (16)^1/2
(25)^1/2 and
5
<
3 + 4
7
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,
1.associative,
(6x - 5x - 1) - 2(x - 3x -and
4) = 6xcommutative
- 5x - 1 - 2x + 6x + 8
Distributive
= 6x - 2x – 5x + 6x – 1+ 8 Associative &
properties.
Commutative
2
2
2
2
2
=
2.
3.
2
4x2 + x + 7
(6x2 - 5x - 1) + 2(x2 - 3x - 4) = 6x2 - 5x - 1 + 2x2 - 6x – 8 Distributive
= 6x2 + 2x2 - 5x - 6x - 1 – 8 Associative &
Commutative
= 8x2 - 11x – 9
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
1
= 6
A1.1.5 Use dimensional (unit) analysis to organize conversions and computations.
1. Convert 5 miles per hour to feet per second.
5 miles * 5280 feet * 1 hour * 1 minute =
1 hour
1 mile
60 minutes 60 seconds
miles to ft
2.
hours to min.
min. to sec.
43 feet = 7.17 ft/sec.
6 sec.
reduce
approximate
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 yard
3.
Convert 30 cents per minute to dollars per hour.
30 cents * 1 dollar * 60 minutes = $18/hour
1 minute
100 cents
1 hour
A1.2.1 Solve linear equations.
Solve the equation:
1.
7a + 2 = 5a - 3a + 8 2.
7a + 2 = 2a + 8
-2a
-2a
7x - 2 + 3x = 4x - 3 + 8
10x – 2 = 4x + 5
-4x
-4x
5a + 2 = 8
-2 -2
6x – 2 = 5
+ 2 +2
5a = 6
5
5
6x = 7
6 6
a = 6/5 or 1.2
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.
2. Solve the inequalities 10 < -3x - 5 <
3. Solve the inequalities -4 < 2x + 6 < 4
28.
-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/
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.
-4 < 2x + 6
and 2x + 6 < 4
-6
-6
-6
-6
-10 < 2x and
2x < -2
2
2
2
2
-5 < x
and
3
so,
-5 > x
and
-5 > x > -11
x > -11
-5 < x < -1
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 ---Solve
3n > 50
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
3.
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.
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.
height in 1
foot
increments
Age in 5 year increments
2. Sketch a reasonable graph for the continuous speed of a school bus’s trip to school.
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.
3. Sketch a reasonable graph for the continuous height off the ground of a person jumping rope.
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
s
p
e
e
d
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)
x
(Try 2 negatives, zero and 2 positive numbers)
(Try 2 negatives, zero and 2 positive numbers)
y
x
y
x
-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
different y – value.
Not a Function, when the x-value equals
1 the y-value equals 1 or –1.
y
Function, each x-value equals
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
2.
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.
x
x = y2
3.
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.
3x = 4y
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.
y
x
y
x
y
x
y
x
y
x
y
0 0
0
0
0 0
0
0
0 0
0
0
1 1
-1,1
1
1 -1,1
1
1
1 ¾
4/
-1 1
no
-1
-1 no
1
-1
-1 -¾
-4/3
domain (x-values) can be all real #’s
range (y-values) are all non-negative #’s
domain (x-values) can be all nonnegative numbers
range (y-values) can be all real #’s
3
1
-1
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.
x-axis
y-axis
x-axis
2.
Draw the graph of the line that passes through
the points (3,2) and (–3,-2)
y-axis
3.
Draw the graph of the line with no slope and passes
through the point (4,0)
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.
4x – 6y = 12
1.
Find the slope and y-intercept of the line 4x - 6y = 12.
-4x
-4x
Easiest solution is to put in slope-intercept form
-6y = -4x – 12
(y=mx + b)
-6
-6
y = 2/3x + 2
m = 2/3 and the y-intercept is 2
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 yintercept is 1. Find the slope using
rise over run. The rise is 1 and the
run is –2. The slope is – ½.
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/
The slope is
3x
-2/
+2
2.
Write the equation of the line
2x - 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
3.
Write the equation of the
line 2x – 3y = -5 in slopeintercept form. What is
the y-intercept of this
line? Explain your
answer.
2x – 3y = -5
-2x
-2x
-3y = -2x – 5
-3
-3
3
the slope is 2 and the y-intercept is
–4.
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.
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.
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.
Find the slope of the first line:
m=(y2 – y1)/(x2 – x1)
m= (10 – 4) / (3 – 1)
m= 6/2 =3
Find the slope of the first line:
m=(y2 – y1)/(x2 – x1)
m= (2 – 3) / (3 – 2)
m = -1 / 1 = -1
Find the slope of the first line:
m=(y2 – y1)/(x2 – x1)
m= (-3 – 2) / (-4 – 1)
m = -5 / -5 = 1
Use slope and the first point to find the yintercept using y = mx + b
Use slope and the first point to find the yintercept using y = mx + b
Use slope and the first point to find the yintercept using y = mx + b
4 = 3*(1) + b
4=3+b
-3 -3
1 = 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/ . Use this slope and the point of (1,4) to
3
find the y-intercept of the perpendicular line.
y = mx + b
4 = (–1/3)(1) + b
+1/3
+1/3
13/ = b
3
Use 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
3 = -1*(2) + b
3 = -2 + b
+2 +2
5 = 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 yaxis the y-intercept is 0.
2 = 1*(1) + b
2=1+b
-1 -1
1 = 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 yintercept is 1
Use 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.
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
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 = x
Since 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
x-axis
y < - ¾ x + 3, the origin is on the graph.
y-axis
2.
Draw the graph of the inequality 2x – 3y >
6
x-axis
solve for y
2x – 3y > 6
-2x
-2x
-3y > -2x + 6
-3
-3
flip inequality
y < 2/3 x – 2, the origin is not on the graph
y-axis
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
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
Lines intersect
+x +x
-2x
-2x
3y = x
6y = -2x + 24
at the point (2,6)
3 3
6
6
y = 1/3 x
y = - 1/3 x + 16
m = 1/3 b = 0
m = - 1/3 b= 4
x-axis
y-axis
x-axis
Lines intersect
at the point (2,1)
y-axis
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
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 Lines intersect at
-x
the point (-5,4)
4y = -2x + 6
y = -x - 1
4
4
y = - ½ x + 3/2
m=- ½
b = 3/2
m=-1
x-axis
b= -1
y-axis
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<
x-axis
5
-x
-x
y < -x + 5
m= -1 and b =5
m=0 and b=4
the origin is on the graph
the origin is on the graph
y-axis
2.
x-axis
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
y-axis
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
-2x
-2x
x
-y < -2x + 3
-1
-1
y < 2x – 3
m = 2 b = -3
x-axis
x+y<6
-x
y < -x + 6
y < -x + 6
m = -1 b
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 = 4x
Since 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
Subtract the 2nd from the 1st
Now solve for y
3x + 4y = 11
-(3x + 2y = 7)
2y = 4
2
2
y = 2,
plug 2 for y into the 1st equation and solve for x
3x + 4y = 11, put 2 in place of y
3x + 4(2) = 11
3x + 8 = 11
-8
-8
3x = 3
3 3
the answer is x = 1 and y = 2
2.
Use addition to solve the equations: 5x – 4y = 2 and 3x + 4y = 14
Align equations vertically
Since the y terms are opposites
Add the 2nd line with the 1st
Now solve for x
plug 2 for x into the 1st equation and solve for y
3.
5x – 4y = 2
3x + 4y = 14
8x = 16
8
8
x=2
the answer is
5x – 4y = 2, replace x with 2
5(2) – 4y = 2
10 – 4y = 2
-10
-10
-4y = -8
-4 -4
y=2
x = 2 and y = 2
Use subtraction to solve the equations: x – 3y = 7 and 2x – 6y = 14
Multiply the 1st equation by 2
2(x - 3y = 7)
Makes x terms alike, now align
Subtract the 2nd from the first
2x – 6y = 14
nd
st
Add the 2 line with the 1
-(2x – 6y = 14)
Now solve for x
0 = 0
Since the left side equals the right side this is an IDENTITY
These 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
Align the equations so the terms match up, then subtract
Plug this answer into the 1st equation
3x + 12y = 48
x + 4y = 16,
y=5
-(3x + 2y = -2)
x + 4(5) = 16
10y = 50
x + 20 = 16
10
10
-20
-20
y = 5
x = -4
the 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
Align the equations so the terms match up, then add
Plug this answer into the 1st equation
3.
6x – 2y = 2
3x – y = 1,
-2x + 2y = -6
4x
= -4
4
4
x = -1
x = -1
3(-1) – y = 1
-3 – y = 1
+3
+3
-y = 4
-1
-1
y = -4
the solution is x = -1 and y = -4
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
-1
2nd equation times 2
Add the equations vertically
Plug the answer into the 1st equation
15x – 6y = -21
8x + 6y = -2
23x
= -23
23
23
x
= -1
5x – 2y = -7,
5(-1) – 2y = -7
-5 - 2y = -7
+5
+5
-2y = -2
-2
-2
y = -1
x=
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 = 1904.76
Plug 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.
2
4x - x2 - 7x - 4x + 2 + 5
Use Commutative and Associative Properties to put like terms
together (watch signs)
2
3x – 11x + 7
Combine like terms and you are finished.
2.
Simplify (5x2 + 3x + 4) + (3x2 - 3x - 6).
5x2 + 3x + 4 + 3x2 - 3x - 6
is in front.
5x2 + 3x2 + 3x - 3x + 4 - 6
(watch signs)
2
8x – 2
3.
Remove parentheses, change all signs inside the parentheses if a minus sign
Use Commutative and Associative Properties to put like terms together
Combine like terms and you are finished..
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.
2
- 5x + 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.
2.
Simplify a2b5/ab2.
Simplify (2a2b4)(-3a3b4c)
a2b5
ab2
Cancel one “a” and two “b”’s
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
3.Simplify (-6x2y3z5)/(-15xy4z2)
-6x2y3z5
factors
-15xy4z2
factors by
Rewrite fraction as a stack. Cancel common
of the coefficients (-3) and reduce variable
subtracting the smallest exponent from the larger
and
leaving the variable in the location of the larger
exponent.
A1.6.3 Find powers and roots of monomials (only when the answer has an integer exponent).
1.
2.
3.
Find the square root of a2b6.
(a2b6)½
ab3
Find the cube of 2x2y3
(2x2y3)3
(2x2y3) (2x2y3) (2x2y3)
8x6y9
Find the cube root of –125a6b9
(–125a6b9)1/3
-5a2 b3
rewrite with the square root
Rewrite by cubing the value (3rd power)
Rewrite without exponents
Multiply number factors, add exponents for variable
Rewrite using the cube root (power of 1/3)
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)
2
4n
+
(-5n)
+
8n
+
(-10)
2
4n + (-5n) + 8n + (-10)
Rewritten in shorten form
2
4n + 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
3
2
12x – 5x + 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)
2
64x
+
24x
+
(-24x)
+
(-9)
2
64x + 24x + (-24x) + (-9)
Rewritten in shorten form
2
64x – 9
Answer in simplified form
A1.6.5 Divide polynomials by monomials.
1.
Divide 4x3y2 + 8xy4 - 6x2y5 by 2xy2.
2x2 +
2xy2
2.
Divide 9a5b3 – 12a4b2 + 3a3b5 by 3a2b2
3a2b2
3.
4y2 - 3 xy3
4x3y2 + 8xy4 - 6x2y5
- 4x3y2
0 + 8xy4
- 8xy4
0 - 6x2y5
+ 6x2y5
0
3a7b - 4a2
+
ab3
This is the answer
Subtract and bring down the next term
What times 2xy2 = 8xy4, 4y2
Subtract and bring down the next term
What times 2xy2 = 6x2y5, -3xy3
Subtract and bring down the next term
No remainder
Set up as a long division problem.
This is your answer
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
3
5
0 + 3a b
What times 3a2b2 = 3a3b5 ,
ab3
3
5
- 3a b
remainder is zero
Divide 8a3b2 + 6a2b3+ 4ab4 by 2ab
3a7b - 4a2
2ab
Set up as a long division problem
+
ab3
Set up as a long division problem.
This is your answer
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
4
0 + 4ab
What times 2ab = 4ab4 , 2b3
4
- 4ab
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 y
There is one x factor in the 1st and 2nd terms and two in the 3rd term. “x” is the common factor
There 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 greatest
Next 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 y
There are 4 x factors in the 1stterm, 3 in the 2nd, and 2 in the 3rd. “x2” is the common factor
There 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 – 25
Reverse 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 – 56
Reverse FOIL, Find F and L first, then make sure O+I = 5x
Choices 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 - 16
x2 + 4x
1.
2.
3.
(x + 4)(x – 4
x(x + 4)
=
x–4
x
This is the answer
This is the answer
3x2 – 4x - 7
x2 - 1
(3x – 7)(x + 1
(x+1)(x-1)
=
3x – 7
x–1
4x3 + 6x2 – 4x
36x2 – 6x - 6
2x(2x2 + 3x – 2)
6 (6x2 – x – 1)
=
2x(2x –1)(x + 2)
6 (3x + 2)(2x – 1)
= x(x + 2)
3(3x + 2)
A1.7.2 Solve algebraic proportions.
To solve these, cross multiply and then solve for x.
Cross multiply:
x + 5 = 3x + 5
1. 4
7
2.
3.
2x + 2
-3
5x - 3
6
=
=
x-2
-2
Cross multiply:
5x - 5
10
Cross multiply:
7(x + 5) = 4(3x + 5)
7x + 35 = 12x + 20
-7x
-7x
35 = 5x + 20
-20
-20
15 = 5x,
x=3
-2(2x + 2) = -3(x – 2)
-4x – 4 = -3x + 6
+4x
+4x
-4 = x + 6
-6
-6
-10 = x
10(5x – 3) = 6(5x – 5)
50x – 30 = 30x – 30
-30x
-30x
20x - 30 = -30
+ 30
+ 30
20x = 0
x=0
Watch your signs