Test 2 – Math 96
Flash Cards
Test 2
Functions: Domain & Range
Systems of Equations
Word Problems – Mixture, investment & DRT
Polynomials: add, subtract, multiply, & divide
Factor
Does the graph represents a function.
State the domain and the range.
Does the graph represents a function.
State the domain and the range.
 Vertical line test - if a vertical line will only
touch the graph at most 1 time – then the
graph represents a function.




Domain: how far left , how far right
Range:
how low, how high
Included use [ ]
Not included use (
)
Does the graph represents a function.
State the domain and the range.
Domain:
Range:
Function?
Does the graph represents a function.
State the domain and the range.
Domain:
(-4, 5]
Range:
(4, -3]
Function?
Yes 
Solve this system of equations by using
the Substitution method.
Solve this system of equations by using
the Substitution method.
 Isolate 1 variable in 1 of the equations;
 substitute that information into the other
equation.
 Solve for the value of the other variable.
 Go back and evaluate with the isolated
variable.
Solve this system of equations by using
the Substitution method.
Solve: x  2 y  3
x  y2
Solve this system of equations by using
the Substitution method.
Solve: x  2 y  3
x  y2
2y  3  y  2
y
-y
y  3  2
+3 +3
y 1
x  2( 1 )  3
x  1
( 1,1 )
Solve this system of equations by using
the Addition/Elimination method.
Solve this system of equations by using
the Addition/Elimination method.
 If necessary, multiply one of both equations
by a number so that when the two equations
are added together one of the variables is
eliminated.
 Solve for the remaining variable; go back to
one of the original equations – substitute in
the value and solve for the other variable.
Solve this system of equations by using
the Addition/Elimination method.
 Solve
2x  3 y  0
5x  2 y  19
Solve this system of equations by using
the Addition/Elimination method.
1 ) multiply by 2: 2x  3 y  0
2 ) multiply by 3: 5x  2 y  19
 3,-2 
4x  6 y  0
2( 3 )  3 y  0
15x  6 y  57
6  3y  0
19x  57
x3
3 y  6
y  2
Solve system of equations by either the
Substitution or Addition method.
Solve system of equations by either the
Substitution or Addition method.
 To use the substitution method if a variable is
isolated.
 To use the addition method when both
variables are on the same side of the
equation.
Investment word problem,
set up a system of equations & solve it.
Investment word problem,
set up a system of equations & solve it.
 1st equation – amounts: what was invested.
 2nd equation – value: sum of percentages
time amount equal the interest.
 Solve by one of the above methods.
Investment word problem,
set up a system of equations & solve it.
 Bret has $50,000 to invest, some at 12% and
the rest at 8%. If his annual interest earned is
$4,740, how much did he invest at each rate?
multiply by  8 
x  y  50000
multiply by 100  .12 x  .08 y  4740
 8 x  8 y  400000
12 x  8 y  474000
4x 
x  $18500 invested at 12%
y  $31500 invested at 8%
74000
Investment word problem,
set up a system of equations & solve it.
 Bret has $50,000 to invest, some at 12% and
the rest at 8%. If his annual interest earned is
$4,740, how much did he invest at each rate?
multiply by  8 
x  y  50000
multiply by 100  .12 x  .08 y  4740
 8 x  8 y  400000
12 x  8 y  474000
4x 
x  $18500 invested at 12%
y  $31500 invested at 8%
74000
For the following mixture word problem,
set up a system of equations and solve it.
For the following mixture word problem,
set up a system of equations and solve it.
 1st equation – amounts: ounces, pounds,
liter, etc.
 2nd equation – value: $ or % multiplied to the
amount added together and equal to the $ or
% multiplied to the total amount.
 Solve by one of the above methods.
For the following mixture word problem,
set up a system of equations and solve it.
Victor mixes a 20% acid solution with a 45% acid solution
to create 100 ml of a 40% acid solution. How many ml of
each solution did he use?
x  y  100
multiply by  20 
.20 x  .45 y  40
multiply by 100 
20 x  20 y  2000
20 x  45 y  4000
25 y  2000
y  80ml of 45% acid
x  20ml of 20% acid
For the following mixture word problem,
set up a system of equations and solve it.
Victor mixes a 20% acid solution with a 45% acid solution
to create 100 ml of a 40% acid solution. How many ml of
each solution did he use?
x  y  100
multiply by  20 
.20 x  .45 y  40
multiply by 100 
20 x  20 y  2000
20 x  45 y  4000
25 y  2000
y  80ml of 45% acid
x  20ml of 20% acid
For the following D = RT word problem,
set up a system of equations and solve it.
For the following D = RT word problem,
set up a system of equations and solve it.
 Label 2 equations:
 Formula: (rate) (time) = distance
mph hour
miles
( t ) and (total – t)

total time

total distance ( d ) and (total – d)
For the following D = RT word problem,
set up a system of equations and solve it.
A sailboat leaves the island traveling at 35 mph. Five
hours later a hydrofoil begins to pursue the sailboat at a
speed of 60 mph. How far from the island will they meet?
For the following D = RT word problem,
set up a system of equations and solve it.
A sailboat leaves the island traveling at 35 mph. Five
hours later a hydrofoil begins to pursue the sailboat at a
speed of 60 mph. How far from the island will they meet?
sailboat :
d  35t
hydrofoil :
d  60  t  5 
35t  60t  300
60t
substitution
d d
d  35 12 
60t
25t  300

1
25
 25t   
d  420 miles
1
25
 300 
t  12
Solve this system of equations
with 3 unknowns.
Solve this system of equations
with 3 unknowns.
 Use the elimination method and 2 of the
original equations to make equation 4.
 Use the other original equation and an
already used original equation to eliminate
the same variable and make equation 5.
 Use equations 4 and 5 to eliminate another
variable; put value back into 4 or 5, put both
values back into and original equation.
Solve this system of equations
with 3 unknowns.
 1) x + y + z = 8
 2) x + 2y – z = 3
 3) 2x – y + z = 3
Solve this system of equations
with 3 unknowns.
 1) x + y + z = 8
 2) x + 2y – z = 3
 3) 2x – y + z = 3
 1) x + y + z = 8
 2) x + 2y – z = 3
 4) 2x + 3y = 11
 2) x + 2y – z = 3
 3) 2x – y + z = 3
 5) 3x + y = 6
4) 2x + 3y = 11
5) -9x – 3y = -18
-7x
= -7
-7
-7
x = 1
2x + 3y = 11
2(1) + 3y = 11
-2
-2
3y = 9
3 3
y=3
Solve for z
x + y+ z=8
(1) + (3) + z = 8
4+z=8
-4
-4
z=4
Add or Subtract the polynomials as
indicated. Simplify your answer.
Add or Subtract the polynomials as
indicated. Simplify your answer.
 Combine like terms −
 Beware of subtraction outside of the
parentheses, change all the signs inside the
parentheses.
Add or Subtract the polynomials as
indicated. Simplify your answer.
x
2
 
 3x  4  2x  x  7
2

Add or Subtract the polynomials as
indicated. Simplify your answer.
x
2
 
 3x  4  2x  x  7
2

x  3x  4  2x  x  7
2
2
 x  4x  11
2
Multiply the following polynomials as
indicated. Simplify your answer.
Multiply the following polynomials as
indicated. Simplify your answer.
 If there is an “outside” exponent, write the
base as many time as indicated by the
exponent.
 Use the distributive property.
 Simplify your answer by combining like terms.
Multiply the following polynomials as
indicated. Simplify your answer.
 2x  5 
 2x  5  2x  5 
2
4x  10x  10x  25
2
4x  20x  25
2
Multiply the following polynomials as
indicated. Simplify your answer.
 2x  5 
 2x  5  2x  5 
2
4x  10x  10x  25
2
4x  20x  25
2
Use long division, where R is less than
the degree of D.
Use long division, where R is less than
the degree of D.
 Divide the leading term of the denominator
into the leading term of the numerator.
 Multiply back to all terms that were in the
denominator.
 Subtract, better yet, add the opposite.
 Repeat above steps till degree “inside” is less
than the degree “outside”.
Use long division, where R is less than
the degree of D.
6 x  5x  1
2x  3
2
3x  2
2x  3 6 x 2  5x  1
 6 x  9x
 4x  1
2

6 x  5x  1
5
 3x  2 
2x  3
2x  3
2
4x  6
5
Use long division, where R is less than
the degree of D.
6 x  5x  1
2x  3
2
3x  2
2x  3 6 x 2  5x  1
 6 x  9x
 4x  1
2

6 x  5x  1
5
 3x  2 
2x  3
2x  3
2
4x  6
5
Factor completely – 2 Terms
Factor completely – 2 Terms




Always look for a common factor

immediately take it out to the front of the expression all
common factors, show what’s left inside ONE set of
parenthesis
Identify the number of terms. If there was a common
factor, then look at the number of terms left inside the
parenthesis.
If there are only two terms, see if it is the sum or
difference of perfect cubes or squares

a2 – b2 = (a + b)(a – b)

a3 – b3 = (a – b)(a2 + ab + b2) OR

a3 + b3 = (a + b)(a2 – ab + b2)
To factor the sum and difference of cubes, write what you
see without exponent 3, “square” “smile” “square”, the
second sign is different, the last sign is always plus.
Factor completely: 40 – 5y3
Factor completely: 40 – 5y3
 Common: 5(8 – y3)
 See exponent 3: 5(23 – y3)
 5(2 – y)(4 + 2y + y2)
Factor completely – 3 Terms
Factor completely – 3 Terms










Always look for a common factor
immediately take it out to the front of the expression all common
factors
show what’s left inside ONE set of parenthesis
Identify the number of terms.
If there are three terms, and the leading coefficient is positive:
find all the factors of the first term, find all the factors of the last term
Within 2 sets of parentheses,

place the factors from the first term in the front of the parentheses

place the factors from the last term in the back of the parentheses
NEVER put common factors together in one parenthesis.
check the last sign,

if the sign is plus: use the SAME signs, the sign of the 2nd term

if the sign is minus: use different signs, one plus and one minus
“smile” to make sure you get the middle term

multiply the inner most terms together then multiply the outer
most terms together, and add the two products together.
Factor completely: 2x2 – 5x – 7
Factor completely: 2x2 – 5x – 7
 Factors of the first term: 1x & 2x
 Factors of the last term:
-1 & 7 or 1 & -7
 (2x – 7)(x + 1)
Factor completely : 4 Terms
Factor completely : 4 Terms

Always look for a common factor
immediately take it out to the front of the expression all
common factors, show what’s left inside ONE set of parenthesis


Identify the number of terms. If there was a common factor, then look
at the number of terms left inside the parenthesis.

Do NOT remove parentheses already given in the expression.

If there are four terms:
try to factor the first three terms if perfect square trinomial



write it as (ax + b)2 – y2
factor as the difference of perfect squares.
try to factor the out a negative from the last three terms



write it as x2 – (ay + b)2
factor as the difference of perfect squares
Factor completely:
x2 – y2 + 10y – 25
Factor completely:
x2 – y2 + 10y – 25
Factor a negative out of the last 3 terms
x2 – (y2 – 10y + 25)
x2 – (y – 5)(y – 5)
x2 – (y – 5)2
(x + (y – 5))(x – (y – 5))
(x + y – 5)(x – y + 5)