final exam review advanced algebra 1. ● solve the system using elimination: m + n = 6 m - n = 5...

Final Exam ReviewAdvanced Algebra 1

● Solve the system using elimination: m + n = 6 m - n = 5

● Notice that the n terms in both equations are additive inverses. So if we add the equations the n terms will cancel.

● So let’s add & solve: m + n = 6 + m - n = 5

2m + 0 = 11

2m = 11

m = 11/2 or 5.5

● Insert the value of m to find n: 5.5 + n = 6

n = .5

● The solution is (5.5, .5).

● Solve the system using elimination: 3s - 2t = 10 4s + t = 6

We could multiply the second equation by 2 and the t terms would be inverses. OR

We could multiply the first equation by 4 and the second equation by -3 to make the s terms inverses.

● Let’s multiply the second equation by 2 to eliminate t. (It’s easier.)

3s - 2t = 10 3s – 2t = 10

2(4s + t = 6) 8s + 2t = 12

● Add and solve: 11s + 0t = 22

11s = 22

s = 2

● Insert the value of s to find the value of t 3(2) - 2t = 10 t = -2

● The solution is (2, -2).

EliminationSolve the system by elimination:

1. -4x + y = -12

4x + 2y = 6

2. 5x + 2y = 12

-6x -2y = -14

3. 5x + 4y = 12

7x - 6y = 40

4. 5m + 2n = -8 4m +3n = 2

Substitution

To solve a system of equations by substitution…

1. Solve one equation for one of the variables.

2. Substitute the value of the variable into the otherequation.

3. Simplify and solve the equation.

4. Substitute back into either equation to find the value of the other variable.

Substitution● Solve the system: x - 2y = -5

y = x + 2

Notice: One equation is already solved for one variable.

Substitute (x + 2) for y in the first equation.

x - 2y = -5

x - 2(x + 2) = -5

● We now have one equation with one variable. Simplify and solve.

x - 2x – 4 = -5

-x - 4 = -5

-x = -1

x = 1

● Substitute 1 for x in either equation to find y.

y = x + 2 y = 1 + 2 so y = 3

● The solution is (1, 3).

Substitution● Let’s check the solution. The answer (1, 3) must check

in both equations.

x - 2y = -5 y = x + 2

1 - 2(3) = -5 3 = 1 + 2

-5 = -5 3 = 3

Substitution● Solve the system: 2p + 3q = 2

p - 3q = -17

● Notice that neither equation is solved for a variable. Since p in the

second equation does not have a coefficient, it will be easier to solve.

p - 3q = -17

p = 3q – 17

● Substitute the value of p into the first equation, and solve.

2p + 3q = 2

2(3q – 17) + 3q = 2

6q – 34 + 3q = 2

9q – 34 = 2

9q = 36

q = 4

Substitution● Substitute the value of q into the second equation to find p.

p = 3q – 17

p = 3(4) – 17

p = -5

● The solution is (-5, 4). (List p first since it comes first alphabetically.)

● Let’s check the solution:

2p + 3q = 2 p – 3q = -17

2(-5) +3(4) = 2 -5 - 3(4) = -17

-10 + 12 = 2 -5 - 12 = -17

2 = 2 -17 = -17

Solve the systems by substitution:

1. x = 4

2x - 3y = -19

2. 3x + y = 7

4x + 2y = 16

3. 2x + y = 5

3x – 3y = 3

4. 2x + 2y = 4

x – 2y = 0

How to Use Elimination Method to Solve System of Equations?

Use elimination to solve each system of equations, if possible. Identify the system as consistent or inconsistent. If the system is consistent, state whether the equations are dependent or independent. Support your results graphically.

a) 3x − y = 7 b) 5x − y = 8 c) x − y = 5

5x + y = 9 −5x + y = −8 x − y = − 2

How to Use Elimination Method to Solve System of Equations? (Cont.)

Solution

a)Eliminate y by addingthe equations.

Find y by substituting x = 2 in either equation.

The solution is (2, −1). The system is consistent and the equations are independent.


If we add the equations we obtain the following result.

The equation 0 = 0 is an identity that is always true. The two equations are equivalent. There are infinitely many solutions.

{(x, y)| 5x − y = 8}

b)


c) If we subtract the second equation fromthe first, we obtain the following result.

The equation 0 = 7 is a contradiction that is never true. Therefore there are no solutions, and the system is inconsistent.

Exponent Rule #1When multiplying two expressions with the

same base you add their exponents.

For example

mn bb mnb

42 xx 42x 6x 222 21 22 212 32 8

Exponent Rule #1

Try it on your own:

mn bb mnb

73.1 hh33.2 2

1073 hh

312 33

27333

Exponent Rule #2When dividing two expressions with the same

base you subtract their exponents.

For example

m

n

b

b mnb

2

4

x

x 24x 2x

Exponent Rule #2

m

n

b

b mnb

2

6

.3h

h

Try it on your own:

3

3.4

3

26h 4h

133 23 9

Exponent Rule #3When raising a power to a power you multiply

the exponents

For example

mnb )( mnb 42 )(x 42x 8x

22 )2( 222 42 16

Exponent Rule #3

Try it on your own

mnb )( mnb 23)(.5 h 23h 6h22 )3(.6 223 43 81

NoteWhen using this rule the exponent can not be

brought in the parenthesis if there is addition or subtraction

222 )2( x 44 2xYou would have to use FOIL in these cases

Exponent Rule #4When a product is raised to a power, each piece

is raised to the power

For example

mab)( mmba2)(xy 22 yx

2)52( 22 52 254 100

Exponent Rule #4

Try it on your own

mab)( mmba

3)(.7 hk 33kh2)32(.8 22 32 94 36

NoteThis rule is for products only. When using this

rule the exponent can not be brought in the parenthesis if there is addition or subtraction

2)2( x 22 2xYou would have to use FOIL in these cases

Exponent Rule #5When a quotient is raised to a power, both the

numerator and denominator are raised to the power

For example

m

b

am

m

b

a

3

y

x3

3

y

x

Exponent Rule #5

Try it on your own

m

b

am

m

b

a

2

.9k

h2

2

k

h

2

2

4.10 2

2

2

4

4

16 4

Zero ExponentWhen anything, except 0, is raised to the zero

power it is 1.

For example

0a 1 ( if a ≠ 0)

0x 1 ( if x ≠ 0)

025 1

Zero Exponent

Try it on your own

0a 1 ( if a ≠ 0)

0.11 h 1 ( if h ≠ 0)

01000.12 100.13 0

Negative ExponentsIf b ≠ 0, then

For example

nb nb

1

2x 2

1

x

23 23

1

9

1

Negative ExponentsIf b ≠ 0, then

Try it on your own:

nb nb

1

3.14 h 3

1

h

32.15 32

1

8

1

Negative Exponents The negative exponent basically flips the part with the negative exponent to the other half of the fraction.

2

1

b

1

2b 2b

2

2

x

1

2 2x 22x

Math MannersFor a problem to be

completely simplified there should not be any negative exponents

Mixed Practice

9

5

3

6.1

d

d 952 d 42 d4

2

d

54 42.2 ee 548 e 98e

Mixed Practice

54.3 q 54q 20q

52.4 lp 5552 pl 5532 pl

Mixed Practice

2

42

)(

)(.5

xy

yx22

48

yx

yx 2428 yx 26 yx

9

253 )(.6

x

xx9

28 )(

x

x

9

16

x

x 916x 7x

Mixed Practice6523246 )()(.7 pnmnm

301218812 pnmnm 301281812 pnm

302030 pnm

Mixed Practice

4

6

)2(

)2(.8

yx

yx

46)2( yx 2)2( yx

)2)(2( yxyx

F2x

Oxy2

Ixy2L

24y22 44 yxyx

Mixed Practice

94

56

.9da

da 9546 da 42 da

4

2

d

a

Find the GCF of each list of numbers.1) 6, 8 and 46

6 = 2 · 3 8 = 2 · 2 · 246 = 2 · 23So the GCF is 2.

2) 144, 256 and 300144 = 2 · 2 · 2 · 3 · 3256 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2300 = 2 · 2 · 3 · 5 · 5So the GCF is 2 · 2 = 4.

Greatest Common Factor

Example

1) x3 and x7

x3 = x · x · x

x7 = x · x · x · x · x · x · x

So the GCF is x · x · x = x3

2) 6x5 and 4x3

6x5 = 2 · 3 · x · x · x

4x3 = 2 · 2 · x · x · x

So the GCF is 2 · x · x · x = 2x3

Find the GCF of each list of terms.


Example

Find the GCF of the following list of terms.

a3b2, a2b5 and a4b7

a3b2 = a · a · a · b · ba2b5 = a · a · b · b · b · b · b a4b7 = a · a · a · a · b · b · b · b · b · b · b

So the GCF is a · a · b · b = a2b2

Notice that the GCF of terms containing variables will use the smallest exponent found amongst the individual terms for each variable.


Example

Factor the polynomial x2 + 13x + 30.Since our two numbers must have a product of 30 and a sum of 13, the two numbers must both be positive.

Positive factors of 30 Sum of Factors

1, 30 31

2, 15 173, 10 13

Note, there are other factors, but once we find a pair that works, we do not have to continue searching.

So x2 + 13x + 30 = (x + 3)(x + 10).

Factoring Polynomials

Example

Factor the polynomial x2 – 11x + 24.Since our two numbers must have a product of 24 and a sum of -11, the two numbers must both be negative.

Negative factors of 24 Sum of Factors

– 1, – 24 – 25

– 2, – 12 – 14 – 3, – 8 – 11

So x2 – 11x + 24 = (x – 3)(x – 8).


Example

Factor the polynomial x2 – 2x – 35.Since our two numbers must have a product of – 35 and a sum of – 2, the two numbers will have to have different signs.

Factors of – 35 Sum of Factors

– 1, 35 34

1, – 35 – 34

– 5, 7 25, – 7 – 2

So x2 – 2x – 35 = (x + 5)(x – 7).


Example

Factor the polynomial x2 – 6x + 10.

Since our two numbers must have a product of 10 and a sum of – 6, the two numbers will have to both be negative.

Negative factors of 10 Sum of Factors

– 1, – 10 – 11

– 2, – 5 – 7Since there is not a factor pair whose sum is – 6, x2 – 6x +10 is not factorable and we call it a prime polynomial.

Prime Polynomials

Example

First multiply the first coefficient by the last number:

Now find the factors that add up to the middle coefficient and break up the middle term:

Now group the first two terms and the last two and GCF each group

Solve by factoring

2 2or

3 3x x

Factor using “difference of two squares.”

9 or 3x x

Solve by factoring

1

2or 3x x

Solve by factoring

Solve by factoring

0 or 4x x

Did you take out GCF?

Solve x2 – 5x = 24.• First write the quadratic equation in standard form.

x2 – 5x – 24 = 0• Now we factor the quadratic using techniques from

the previous sections.

x2 – 5x – 24 = (x – 8)(x + 3) = 0• We set each factor equal to 0.

x – 8 = 0 or x + 3 = 0, which will simplify to

x = 8 or x = – 3

Solving Quadratic Equations

Example

Continued.

Creating a Perfect Square Trinomial

In the following perfect square trinomial, the constant term is missing. X2 + 14x + ____

Find the constant term by squaring half the coefficient of the linear term.

(14/2)2

X2 + 14x + 49

Perfect Square Trinomials

Create perfect square trinomials.

x2 + 20x + ___

x2 - 4x + ___

x2 + 5x + ___

100

4

25/4

Solving Quadratic Equations by Completing

the SquareSolve the following

equation by completing the square:

Step 1: Move quadratic term, and linear term to left side of the equation

2 8 20 0x x

2 8 20x x


the Square

Step 2: Find the term that completes the square on the left side of the equation. Add that term to both sides.

2 8 =20 + x x 21

( ) 4 then square it, 4 162

8

2 8 2016 16x x

Solving Quadratic Equations by Completing the Square

Step 3: Factor the perfect square trinomial on the left side of the equation. Simplify the right side of the equation.

2 8 2016 16x x

2

( 4)( 4) 36

( 4) 36

x x

x


the Square

Step 4: Take the square root of each side

2( 4) 36x

( 4) 6x


the Square

Step 5: Set up the two possibilities and solve

4 6

4 6 an

d 4 6

10 and 2 x=

x

x x

x

Solve by Completing the Square

2 6 16 0x x 2 6 16x x

+9 +92 6 9 25x x

23 25x

3 5x 3 5x 8x 2x


2 22 21 0x x 2 22 21x x

+121 +1212 22 121 100x x

211 100x

11 10x 11 10x 21x 1x


2 2 5 0x x 2 2 5x x

+1 +12 2 1 6x x

21 6x

1 6x 1 6x

Example 1: Graphing a Quadratic Function Graph y = 3x2 – 6x + 1.

Step 1 Find the axis of symmetry.

= 1

The axis of symmetry is x = 1. Simplify.

Use x = . Substitute 3

for a and –6 for b.

Step 2 Find the vertex.y = 3x2 – 6x + 1 = 3(1)2 – 6(1) + 1= 3 – 6 + 1= –2

The vertex is (1, –2).

The x-coordinate of the vertex is 1. Substitute 1 for x.

Simplify.The y-coordinate is –2.

Example 1 Continued

Step 3 Find the y-intercept.

y = 3x2 – 6x + 1

y = 3x2 – 6x + 1

The y-intercept is 1; the graph passes through (0, 1).

Identify c.

Check It Out! Example 1a

Graph the quadratic function.

y = 2x2 + 6x + 2

Step 1 Find the axis of symmetry.

Simplify.

Use x = . Substitute 2

for a and 6 for b.

The axis of symmetry is x .

Step 2 Find the vertex.

y = 2x2 + 6x + 2

Simplify.

Check It Out! Example 1a Continued

= 4 – 9 + 2

= –2

The x-coordinate of the vertex is

. Substitute for

x.

The y-coordinate is .

The vertex is .

Ex: Write the first 4 terms of this sequence with:

First term: a1 = 7

Common ratio = 1/3

an = a1 * r n-1

Now find the first five terms:a1 = 7(1/3) (1-1) = 7a2 = 7(1/3) (2-1) = 7/3a3 = 7(1/3) (3-1) = 7/9a4 = 7(1/3) (4-1) = 7/27a5 = 7(1/3) (5-1) = 7/81

Geometric Sequence ProblemFind the 19th term in the sequence of

11,33,99,297 . . .

a19 = 11(3)18 =4,261,626,379

Common ratio = 3

a19 = 11 (3) (19-1)

Start with the sequence formula

Find the common ratio between the values.

Plug in known values

Simplify

an = a1 * r n-1

Let’s try oneFind the 10th term in the sequence of

1, -6, 36, -216 . . .

a10 = 1(-6)9 = -10,077,696

Common ratio = -6

a10 = 1 (-6) (10-1)

Start with the sequence formula

Find the common ratio between the values.

Plug in known values

Simplify

an = a1 * r n-1

Exponential Growth & Decay ModelsGrowth: a(1 + r)x

Decay: a(1 – r)x

a = initial amount before measuring growth/decayr = growth/decay rate (often a percent)x = number of time intervals that have passed

Example #1The value of an iPad

decreases at 35% per year. If the starting price of the iPad is $500, write the exponential function.

How much will your iPad be worth after 5 years?

$58.01

Example #2Suppose the acreage of forest is decreasing by 2%

per year because of development. If there are currently 4,500,000 acres of forest, determine the amount of forest land after 5 years.

Example #3A bank account balance, A, for an account starting

with P dollars, earning an interest rate, r, and left untouched for t years, compounded n times per year, can be calculated as

Find a bank account balance to the nearest dollar, if the account starts with $100, has a rate of 4%, and the money is left in the account for 12 years compounded monthly.

Given the original principal, the annual interest rate, and the amount of time for each investment, and the type of compounded interest, find the amount at the end of the investment.

1.) P = $1,250; r = 8.5%; t = 3 years; quarterly

2.) P = $2,575; r = 6.25%; t = 5 years; annually

final exam review advanced algebra 1. ● solve the system using elimination: m + n = 6 m - n = 5...

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