Final Exam Review Advanced 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).

Elimination Solve the system by elimination: 1. -4x + y = -12 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 other equation. 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

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 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 adding the equations. Find y by substituting x = 2 in either equation. The solution is (2, −1). The system is consistent and the equations are independent.

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

How to Use Elimination Method to Solve System of Equations? (Cont.) If we subtract the second equation from the 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 #1 When multiplying two expressions with the same base you add their exponents. For example

Exponent Rule #1 Try it on your own:

Exponent Rule #2 When dividing two expressions with the same base you subtract their exponents. For example

Exponent Rule #2 Try it on your own:

Exponent Rule #3 When raising a power to a power you multiply the exponents For example

Exponent Rule #3 Try it on your own

Note When using this rule the exponent can not be brought in the parenthesis if there is addition or subtraction You would have to use FOIL in these cases

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

Exponent Rule #4 Try it on your own

Note This rule is for products only. When using this rule the exponent can not be brought in the parenthesis if there is addition or subtraction You would have to use FOIL in these cases

Exponent Rule #5 When a quotient is raised to a power, both the numerator and denominator are raised to the power For example

Exponent Rule #5 Try it on your own

Zero Exponent ( if a ≠ 0) ( if x ≠ 0) When anything, except 0, is raised to the zero power it is 1. For example ( if a ≠ 0) ( if x ≠ 0)

Zero Exponent Try it on your own ( if a ≠ 0) ( if h ≠ 0)

Negative Exponents If b ≠ 0, then For example

Negative Exponents If b ≠ 0, then Try it on your own:

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

Math Manners For a problem to be completely simplified there should not be any negative exponents

Mixed Practice

Mixed Practice

Mixed Practice

Mixed Practice

Mixed Practice F O I L

Mixed Practice

Greatest Common Factor Example Find the GCF of each list of numbers. 6, 8 and 46 6 = 2 · 3 8 = 2 · 2 · 2 46 = 2 · 23 So the GCF is 2. 144, 256 and 300 144 = 2 · 2 · 2 · 3 · 3 256 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 300 = 2 · 2 · 3 · 5 · 5 So the GCF is 2 · 2 = 4.

Greatest Common Factor Example Find the GCF of each list of terms. x3 and x7 x3 = x · x · x x7 = x · x · x · x · x · x · x So the GCF is x · x · x = x3 6x5 and 4x3 6x5 = 2 · 3 · x · x · x 4x3 = 2 · 2 · x · x · x So the GCF is 2 · x · x · x = 2x3

Greatest Common Factor Example Find the GCF of the following list of terms. a3b2, a2b5 and a4b7 a3b2 = a · a · a · b · b a2b5 = 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.

Factoring Polynomials 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 17 3, 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).

Factoring Polynomials 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 2 5, – 7 – 2 So x2 – 2x – 35 = (x + 5)(x – 7).

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

First multiply the first coefficient by the last number:   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      

First multiply the first coefficient by the last number:   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      

First multiply the first coefficient by the last number:   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      

First multiply the first coefficient by the last number:   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      

First multiply the first coefficient by the last number:   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      

First multiply the first coefficient by the last number:   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      

Factor using “difference of two squares.” Solve by factoring Factor using “difference of two squares.”

Solve by factoring

Solve by factoring

Solve by factoring Did you take out GCF?

Solving Quadratic Equations Example 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 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 Square Solve the following equation by completing the square: Step 1: Move quadratic term, and linear term to left side of the equation

Solving Quadratic Equations by Completing the Square Step 2: Find the term that completes the square on the left side of the equation. Add that term to both sides.

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.

Solving Quadratic Equations by Completing the Square Step 4: Take the square root of each side

Solving Quadratic Equations by Completing the Square Step 5: Set up the two possibilities and solve

Solve by Completing the Square +9

Solve by Completing the Square +121

Solve by Completing the Square +1

Example 1: Graphing a Quadratic Function Graph y = 3x2 – 6x + 1. Step 1 Find the axis of symmetry. Use x = . Substitute 3 for a and –6 for b. = 1 Simplify. The axis of symmetry is x = 1. Step 2 Find the vertex. y = 3x2 – 6x + 1 The x-coordinate of the vertex is 1. Substitute 1 for x. = 3(1)2 – 6(1) + 1 = 3 – 6 + 1 Simplify. = –2 The y-coordinate is –2. The vertex is (1, –2).

Step 3 Find the y-intercept. Example 1 Continued Step 3 Find the y-intercept. y = 3x2 – 6x + 1 y = 3x2 – 6x + 1 Identify c. The y-intercept is 1; the graph passes through (0, 1).

Check It Out! Example 1a Graph the quadratic function. y = 2x2 + 6x + 2 Step 1 Find the axis of symmetry. Use x = . Substitute 2 for a and 6 for b. Simplify. The axis of symmetry is x .

Check It Out! Example 1a Continued Step 2 Find the vertex. y = 2x2 + 6x + 2 The x-coordinate of the vertex is . Substitute for x. = 4 – 9 + 2 Simplify. The y-coordinate is . = –2 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 a1 = 7(1/3) (1-1) = 7 a2 = 7(1/3) (2-1) = 7/3 a3 = 7(1/3) (3-1) = 7/9 a4 = 7(1/3) (4-1) = 7/27 a5 = 7(1/3) (5-1) = 7/81 Now find the first five terms:

Geometric Sequence Problem Find the 19th term in the sequence of 11,33,99,297 . . . an = a1 * r n-1 Start with the sequence formula Find the common ratio between the values. Common ratio = 3 a19 = 11 (3) (19-1) Plug in known values a19 = 11(3)18 =4,261,626,379 Simplify

Let’s try one an = a1 * r n-1 Common ratio = -6 a10 = 1 (-6) (10-1) Find the 10th term in the sequence of 1, -6, 36, -216 . . . an = a1 * r n-1 Start with the sequence formula Find the common ratio between the values. Common ratio = -6 a10 = 1 (-6) (10-1) Plug in known values a10 = 1(-6)9 = -10,077,696 Simplify

Exponential Growth & Decay Models Growth: a(1 + r)x Decay: a(1 – r)x a = initial amount before measuring growth/decay r = growth/decay rate (often a percent) x = number of time intervals that have passed

Example #1 The 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 #2 Suppose 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 #3 A 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