Algebra 1 Final Exam Review – 5 days (2nd Semester)

Day 1

Solve an Inequality - 5 -5 w < 3 - 5 -5 w < 3 All numbers less than 3 are solutions to this problem! 5 10 15 -20 -15 -10 -5 -25 20 25

More Examples -8 -8 r -10 All numbers greater than-10 (including -10) ≥ 5 10 15 -20 -15 -10 -5 -25 20 25

More Examples 2 2 x > -1 All numbers greater than -1 make this problem true! 5 10 15 -20 -15 -10 -5 -25 20 25

All numbers less than 8 (including 8) More Examples -8 -8 2h ≤ 16 2 2 h ≤ 8 All numbers less than 8 (including 8) 5 10 15 -20 -15 -10 -5 -25 20 25

x + 3 > -4 6d > 24 2x - 8 < 14 -2c – 4 < 2 Solve the inequality on your own. x + 3 > -4 6d > 24 2x - 8 < 14 -2c – 4 < 2 x > -7 d ≥ 4 x < 11 c ≥ -3

Any time you multiply or divide both sides of an inequality by a NEGATIVE, you must REVERSE THE SIGN!!!! TRY SOLVING THIS:

The solution would look like this:

SOLVE THIS:

SOLVE THIS:

SOLVE THIS:

SOLVE THIS:

Solve the inequality and graph the solutions. y ≤ 4y + 18 To collect the variable terms on one side, subtract y from both sides. y ≤ 4y + 18 –y –y 0 ≤ 3y + 18 Since 18 is added to 3y, subtract 18 from both sides to undo the addition. –18 – 18 –18 ≤ 3y Since y is multiplied by 3, divide both sides by 3 to undo the multiplication. y  –6

Solve the inequality and graph the solutions. 4m – 3 < 2m + 6 To collect the variable terms on one side, subtract 2m from both sides. –2m – 2m 2m – 3 < + 6 Since 3 is subtracted from 2m, add 3 to both sides to undo the subtraction. + 3 + 3 2m < 9 Since m is multiplied by 2, divide both sides by 2 to undo the multiplication.

Solve the inequality and graph the solutions. Check your answer. 4x ≥ 7x + 6 –7x –7x To collect the variable terms on one side, subtract 7x from both sides. –3x ≥ 6 Since x is multiplied by –3, divide both sides by –3 to undo the multiplication. Change ≥ to ≤. x ≤ –2 The solution set is {x:x ≤ –2}. –10 –8 –6 –4 –2 2 4 6 8 10

Solve the inequality and graph the solutions. Check your answer. 5t + 1 < –2t – 6 5t + 1 < –2t – 6 +2t +2t 7t + 1 < –6 To collect the variable terms on one side, add 2t to both sides. Since 1 is added to 7t, subtract 1 from both sides to undo the addition. – 1 < –1 7t < –7 Since t is multiplied by 7, divide both sides by 7 to undo the multiplication. 7t < –7 7 7 t < –1 The solution set is {t:t < –1}. –5 –4 –3 –2 –1 1 2 3 4 5

Solve the inequality and graph the solutions. 2(k – 3) > 6 + 3k – 3 Distribute 2 on the left side of the inequality. 2(k – 3) > 3 + 3k 2k – 6 > 3 + 3k –2k – 2k To collect the variable terms, subtract 2k from both sides. –6 > 3 + k Since 3 is added to k, subtract 3 from both sides to undo the addition. –3 –3 –9 > k k < -9

Day 2

Solving Compound Inequalities

Multiple Choice Solve: +6 +6 +6

Multiple Choice Solve

Solving Compound Inequalities At the end you must flip the whole inequality to have all the signs point to the left and lower numbers on the left

Solve At the end you must flip the whole inequality to have all the signs point to the left and lower numbers on the left

Solve for 3x + 2 < 14 and 2x – 5 > –11 3x + 2 < 14 2x – 5 > -11 -2 -2 +5 +5 3x < 12 2x > -6 x < 4 AND x > -3

Solve “Tree it up”: 5w + 3 = 7 OR 5w + 3 = -7 Solve both equations for w 5w + 3 = 7 5w + 3 = -7 5w = 4 5w = -10 w = w = -2

The expression x  2 can be equal to 5 or 5. SOLUTION Solve | x  2 |  5 Solve | x  2 |  5 The expression x  2 can be equal to 5 or 5. SOLUTION The expression x  2 can be equal to 5 or 5. x  2 IS POSITIVE | x  2 |  5 x  2  5 x  7 x  2 IS POSITIVE | x  2 |  5 x  2  5 x  7 x  3 x  2 IS NEGATIVE | x  2 |  5 x  2  5 x  2 IS POSITIVE x  2 IS POSITIVE x  2  5 x  2 IS NEGATIVE x  2 IS NEGATIVE x  2  5 | x  2 |  5 | x  2 |  5 x  2  5 x  2  5 x  7 x  3 The equation has two solutions: 7 and –3. CHECK | 7  2 |  | 5 |  5 | 3  2 |  | 5 |  5

Isolate the absolute value expression on one side of the equation. Solve | 2x  7 |  5  4 Isolate the absolute value expression on one side of the equation. SOLUTION 2x  7 IS POSITIVE 2x  7  9 2x  7 IS NEGATIVE 2x  2 | 2x  7 |  -9 2x  7  9 2x  16 x = 8 x  1 The equation has two solutions: 8 and –1.

-5 -5 x = 1 x = -1 2x = 2 2x = -2 The solutions are -1 and 1 Solve -5 -5 Subtract 5 from both sides “TREE IT UP” 2x = 2 2x = -2 x = 1 x = -1 The solutions are -1 and 1

“greatOR” “less thAND” This can be written as 1  x  7. Solve | x  4 | < 3 x  4 IS POSITIVE x  4 IS NEGATIVE x  4  3 x  4  3 x  7 x  1 Reverse inequality symbol!!! The solution is all real numbers greater than 1 and less than 7. This can be written as 1  x  7.

x ≥ 4 OR x ≤ -5 Solve | 2x  1 |  3  6 +3 +3 2x + 1 IS POSITIVE +3 +3 2x + 1 IS POSITIVE 2x + 1 IS NEGATIVE 2x  1  9 2x  1  9 2x  8 2x  10 x  4 x  5 x ≥ 4 OR x ≤ -5

Examples or or Check and verify on a number line. Numbers above 6 or below -1 keep the absolute value greater than 7. Numbers between them make the absolute value less than 7.

Solve absolute-value inequalities. Solve |x – 4|  5. Case 1: Case 2: x – 4 is positive x – 4 is negative x – 4  5 x – 4  –5 x  9 x  –1 solution: –1  x  9

Solve absolute-value inequalities. Solve |4x – 2|  -18. Exception alert!!!! When the absolute value equals a negative value, there is no solution.

TRY THIS Solve absolute-value inequalities. Solve |2x – 6|  18. Case 2: Case 1: 2x – 6 is negative 2x – 6 is positive 2x – 6  18 2x - 6  –18 2x  24 2x  –12 x  12 x  –6 Solution: –6  x  12

TRY THIS Solve absolute-value inequalities. Solve |3x – 2|  -4. Exception alert!!!! When the absolute value equals a negative value, there is no solution.

Day 3

Solve: by ELIMINATION x + y = 12 -x + 3y = -8 Like variables must be lined under each other. 4y = 4 We need to eliminate (get rid of) a variable. The x’s will be the easiest. So, we will add the two equations. Divide by 4 y = 1 Then plug in the one value to find the other: ANSWER: (11, 1)

x + y =12 11 + 1 = 12 12 = 12 -x + 3y = -8 -11 + 3(1) = -8 -11 + 3 = -8 -8 = -8

Solve: by ELIMINATION 5x - 4y = -21 -2x + 4y = 18 Like variables must be lined under each other. 3x = -3 We need to eliminate (get rid of) a variable. The y’s be will the easiest.So, we will add the two equations. Divide by 3 x = -1 Then plug in the one value to find the other: ANSWER: (-1, 4)

5x - 4y = -21 5(-1) – 4(4) = -21 -5 - 16 = -21 -21 = -21 -2(-1) + 4(4) = 18 2 + 16 = 18 18 = 18

Solve: by ELIMINATION 2x + 7y = 31 5x - 7y = - 45 Like variables must be lined under each other. 7x = -14 We need to eliminate (get rid of) a variable. The y’s will be the easiest. So, we will add the two equations. Divide by 7 x = -2 Then plug in the one value to find the other: ANSWER: (-2, 5)

2x + 7y = 31 2(-2) + 7(5) = 31 -4 + 35 = 31 31 = 31 5x – 7y = - 45 5(-2) - 7(5) = - 45 -10 - 35 = - 45 - 45 =- 45

Solve: by ELIMINATION x + y = 30 x + 7y = 6 Like variables must be lined under each other. Solve: by ELIMINATION x + y = 30 x + 7y = 6 We need to eliminate (get rid of) a variable. To simply add this time will not eliminate a variable. If one of the x’s was negative, it would be eliminated when we add. So we will multiply one equation by a – 1.

x + y = 30 x + y = 30 -x – 7y = -6 ( x + 7y = 6 ) -1 -6y = 24 -6 -6 Now add the two equations and solve. -6 -6 Then plug in the one value to find the other: y = - 4 ANSWER: (34, -4)

x + y = 30 34 + - 4 = 30 30 = 30 x + 7y = 6 34 + 7(- 4) = 6 34 - 28 = 6 6 = 6

Solve: by ELIMINATION x + y = 4 2x + 3y = 9 Like variables must be lined under each other. Solve: by ELIMINATION x + y = 4 2x + 3y = 9 We need to eliminate (get rid of) a variable. To simply add this time will not eliminate a variable. If there was a –2x in the 1st equation, the x’s would be eliminated when we add. So we will multiply the 1st equation by a – 2.

( ) -2 y = 1 x + y = 4 -2x – 2y = - 8 2x + 3y = 9 2x + 3y = 9 ANSWER: Now add the two equations and solve. Then plug in the one value to find the other: ANSWER: (3, 1)

x + y = 4 3 + 1 = 4 4 = 4 2x + 3y = 9 2(3) + 3(1) = 9 6 + 3 = 9 9 = 9

3  3  3  3 = 81 1. Evaluate the following exponential expressions: B. 34 = C. 23 = D. (-1) = 4  4 = 16 3  3  3  3 = 81 2  2  2 = 8 7 -1  -1  -1  -1  -1  -1  -1 = -1 REMEMBER TO PUT PARENTHESES AROUND NEGATIVE NUMBERS WHEN USING YOUR CALCULATOR!!!!!

Laws of Exponents

Zero Exponents = 1 a A nonzero based raise to a zero exponent is equal to one

) ( a Negative Exponents 1 -n a ______ = n a A nonzero base raised to a negative exponent is the reciprocal of the base raised to the positive exponent.

Basic Examples

Basic Examples

Basic Examples

Examples 1. 2. 3. 4.

Example 7.2  106 Write 7,200,000 in scientific notation Big Number means Positive Exponent 7.2  106

Example 4.76  102 Write 476 in scientific notation. Big Number means Positive Exponent 4.76  102

Example 6.2  10-3 Write 0.0062 in scientific notation. Small Number means Negative Exponent 6.2  10-3

Example 1. Write these numbers in standard notation: a.) 4.6 x 10ˉ³ b.) 4.6 x 10 2. Saturn is about 875,000,000 miles from the sun. What is this distance in scientific notation? 4.6 0.0046 6 4.6 4,600,000 8.75  108

Day 4

Simplify = ±2 = ±4 = ±5 This is a piece of cake! = ±10 = ±12

Simplify = = = = = = = = = = Perfect Square Factor * Other Factor LEAVE IN RADICAL FORM = = = = = =

Simplify = = = = = = = = = = Perfect Square Factor * Other Factor LEAVE IN RADICAL FORM = = = = = =

Write original equation. Solve x 2 = 8 algebraically. 1 2 SOLUTION 1 2 x 2 = 8 Write original equation. 2   2 x 2 = 16 Multiply each side by 2. x =  4 Find the square root of each side.

Substitute 13 and –13 into the original equation. Solve using square roots. Check your answer. x2 = 169 Solve for x by taking the square root of both sides. Use ± to show both square roots. x = ± 13 The solutions are 13 and –13. Substitute 13 and –13 into the original equation. Check x2 = 169 (13)2 169 169 169  x2 = 169 (–13)2 169 169 169 

Solve using square roots. x2 = –49 There is no real number whose square is negative. Answer: There is no real solution.

Substitute 11 and –11 into the original equation. Solve using square roots. Check your answer. x2 = 121 Solve for x by taking the square root of both sides. Use ± to show both square roots. x = ± 11 The solutions are 11 and –11. Substitute 11 and –11 into the original equation. Check x2 = 121 (11)2 121 121 121  x2 = 121 (–11)2 121 121 121 

Substitute 0 into the original equation. Solve using square roots. Check your answer. Solve for x by taking the square root of both sides. Use ± to show both square roots. x2 = 0 x = 0 The solution is 0. Check x2 = 0 (0)2 0 0 0  Substitute 0 into the original equation.

Solve using square roots. Check your answer. x2 = –16 There is no real number whose square is negative. There is no real solution.

Solve using square roots. x2 + 7 = 7 –7 –7 x2 + 7 = 7 x2 = 0 Subtract 7 from both sides. Take the square root of both sides. The solution is 0.

Solve using square roots. 16x2 – 49 = 0 Add 49 to both sides. 16x2 – 49 = 0 +49 +49 Divide by 16 on both sides. Take the square root of both sides. Use ± to show both square roots.

Solve using the quadratic formula.

Solve using the quadratic formula

Solve using the quadratic formula

1. Add the following polynomials: (9y - 7x + 15a) + (-3y + 8x - 8a) Group your like terms. 9y - 3y - 7x + 8x + 15a - 8a 6y + x + 7a

2. Add the following polynomials: (3a2 + 3ab - b2) + (4ab + 6b2) Combine your like terms. 3a2 + 3ab + 4ab - b2 + 6b2 3a2 + 7ab + 5b2

Just combine like terms. 3. Add the following polynomials using column form: (4x2 - 2xy + 3y2) + (-3x2 - xy + 2y2) Just combine like terms. x2 - 3xy + 5y2

You need to distribute the negative!! 4. Subtract the following polynomials: (9y - 7x + 15a) - (-3y + 8x - 8a) You need to distribute the negative!! (9y - 7x + 15a) + (+ 3y - 8x + 8a) Group the like terms. 9y + 3y - 7x - 8x + 15a + 8a 12y - 15x + 23a

5. Subtract the following polynomials: (7a - 10b) - (3a + 4b) Distribute the negative (7a - 10b) + (- 3a - 4b) Group the like terms. 7a - 3a - 10b - 4b 4a - 14b

6. Subtract the following: (4x2 - 2xy + 3y2) - (-3x2 – xy + 2y2) Distribute the negative!!! 4x2 – 2xy + 3y2 + 3x2 + xy – 2y2 7x2 - xy + y2

Find the sum or difference. (5a – 3b) + (2a + 6b)

Find the sum or difference. (5a – 3b) – (2a + 6b)

Multiply (y + 4)(y – 3) y2 + y – 12 y2 – y – 12 y2 + 7y – 12

Multiply (2a – 3b)(2a + 4b) 4a2 + 14ab – 12b2 4a2 – 14ab – 12b2 4a2 + 8ab – 6ba – 12b2 4a2 + 2ab – 12b2 4a2 – 2ab – 12b2

Group and combine like terms. 5) Multiply (2x - 5)(x2 - 5x + 4) You cannot use FOIL because they are not BOTH binomials. You must use the distributive property. 2x(x2 - 5x + 4) - 5(x2 - 5x + 4) 2x3 - 10x2 + 8x - 5x2 + 25x - 20 Group and combine like terms. 2x3 - 10x2 - 5x2 + 8x + 25x - 20 2x3 - 15x2 + 33x - 20

Multiply (2p + 1)(p2 – 3p + 4) 2p3 + 2p3 + p + 4 y2 – y – 12

Multiply each: = 2x2 + 9x + -5 = 6w2 + -19w + 10

Multiply each: 4a4 + 2a3 – 2a2 + 2a2 + a – 1 4a4 + 2a3 + a - 1 Distribute the binomial 4a4 + 2a3 – 2a2 + 2a2 + a – 1 4a4 + 2a3 + a - 1

Use the FOIL method to multiply these binomials: Multiply each: Use the FOIL method to multiply these binomials: (3a + 4)(2a + 1) (x + 4)(x – 5) (x + 5)(x – 5) (c - 3)(2c - 5) 5) (2w + 3)(2w – 3) = 6a2 + 3a + 8a + 4 = 6a2 + 11a + 4 = x2 - 1x - 20 = x2 - 5x + 4x - 20 = x2 - 25 = x2 - 5x + 5x - 25 = 2c2 - 5c - 6c + 15 = 2c2 - 11c + 15 = 4w2 - 6w + 6w - 9 = 4w2 - 9

Day 5

Review: What is the GCF of 25a2 and 15a? Let’s go one step further… 1) FACTOR 25a2 + 15a. Find the GCF and divide each term 25a2 + 15a = 5a( ___ + ___ ) Check your answer by distributing. 5a 3

2) Factor 18x2 - 12x3. Find the GCF 6x2 Divide each term by the GCF 18x2 - 12x3 = 6x2( ___ - ___ ) Check your answer by distributing. 3 2x

3) Factor 28a2b + 56abc2. GCF = 28ab Divide each term by the GCF 28a2b + 56abc2 = 28ab ( ___ + ___ ) Check your answer by distributing. 28ab(a + 2c2) a 2c2

Factor 20x2 - 24xy x(20 – 24y) 2x(10x – 12y) 4(5x2 – 6xy) 4x(5x – 6y)

5) Factor 28a2 + 21b - 35b2c2 GCF = 7 Divide each term by the GCF Check your answer by distributing. 7(4a2 + 3b – 5b2c2) 4a2 3b 5b2c2

Factor 16xy2 - 24y2z + 40y2 2y2(8x – 12z + 20) 4y2(4x – 6z + 10) 8xy2z(2 – 3 + 5)

Factor These Trinomials! Factor each trinomial, if possible. The first four do NOT have leading coefficients, the last two DO have leading coefficients. Watch out for signs!! 1) t2 – 4t – 21 2) x2 + 12x + 32 3) x2 –10x + 24 4) x2 + 3x – 18 5) 2x2 + x – 21 6) 3x2 + 11x + 10

Solution #1: t2 – 4t – 21 t2 – 4t – 21 = (t + 3)(t - 7) 1) Factors of -21: 1 • -21, -1 • 21 3 • -7, -3 • 7 2) Which pair adds to (- 4)? 3) Write the factors. t2 – 4t – 21 = (t + 3)(t - 7)

Solution #2: x2 + 12x + 32 x2 + 12x + 32 = (x + 4)(x + 8) 1) Factors of 32: 1 • 32 2 • 16 4 • 8 2) Which pair adds to 12 ? 3) Write the factors. x2 + 12x + 32 = (x + 4)(x + 8)

Solution #3: x2 - 10x + 24 x2 - 10x + 24 = (x - 4)(x - 6) 1) Factors of 32: 1 • 24 2 • 12 3 • 8 4 • 6 -1 • -24 -2 • -12 -3 • -8 -4 • -6 2) Which pair adds to -10 ? None of them adds to (-10). For the numbers to multiply to +24 and add to -10, they must both be negative! 3) Write the factors. x2 - 10x + 24 = (x - 4)(x - 6)

Solution #4: x2 + 3x - 18 x2 + 3x - 18 = (x - 3)(x + 18) 1) Factors of -18: 1 • -18, -1 • 18 2 • -9, -2 • 9 3 • -6, -3 • 6 2) Which pair adds to 3 ? 3) Write the factors. x2 + 3x - 18 = (x - 3)(x + 18)

Solution #5: 2x2 + x - 21 2x2 + x - 21 = (x - 3)(2x + 7) 3 1) Multiply 2 • (-21) = - 42; list factors of - 42. 1 • -42, -1 • 42 2 • -21, -2 • 21 3 • -14, -3 • 14 6 • -7, -6 • 7 2) Which pair adds to 1 ? 3) Write the temporary factors. ( x - 6)( x + 7) 2 2 4) Put “2” underneath. 3 ( x - 6)( x + 7) 2 5) Reduce (if possible). 6) Move denominator(s)in front of “x”. ( x - 3)( 2x + 7) 2x2 + x - 21 = (x - 3)(2x + 7)

Solution #6: 3x2 + 11x + 10 3x2 + 11x + 10 = (3x + 5)(x + 2) 2 1) Multiply 3 • 10 = 30; list factors of 30. 1 • 30 2 • 15 3 • 10 5 • 6 2) Which pair adds to 11 ? 3) Write the temporary factors. ( x + 5)( x + 6) 3 3 4) Put “3” underneath. 2 ( x + 5)( x + 6) 3 5) Reduce (if possible). 6) Move denominator(s)in front of “x”. ( 3x + 5)( x + 2) 3x2 + 11x + 10 = (3x + 5)(x + 2)

Set up and solve this proportion. 8) 5 is to 4 as 45 is to x. 5 Now cross multiply 45 ____ = ____ 4 x 5x = 180 5 5 x = 36

First - Set up & solve the proportion. 9) 3 is to 2 as x is to 18. 3 Now cross multiply x ____ = ____ 2 18 2x = 54 2 2 x = 27

NO set up this time!! Just solve….. 10) 6x = 144 6 6 x = 24

You try it!! 11) Now Reduce 16m = 8 16 16 m =

Use cross products to solve the proportion. 25 20 45 t 1. = t = 36 2. x 9 = 19 57 x = 3 2 3 r 36 3. = r = 24 n 10 28 8 4. = n = 35

Problem 1 Factor the numerator and denominator Divide out the common factors. Write in simplified form.

Problem 2 Factor the numerator and denominator You Try It Problem 2 Factor the numerator and denominator Divide out the common factors. Write in simplified form.

Problem 3 Factor the numerator and denominator You Try It Problem 3 Factor the numerator and denominator Divide out the common factors. Write in simplified form.

Problem 4 Factor the numerator and denominator You Try It Problem 4 Factor the numerator and denominator Divide out the common factors. Write in simplified form.

Problem 5 Factor the numerator and denominator You Try It Problem 5 Factor the numerator and denominator Divide out the common factors. Write in simplified form.

YOU TRY IT What are the excluded values of the variables for the following rational expressions?

Problem 1 Solution y  0 z  0

Problem 2 Solution 2x - 12 = 0 2x - 12 + 12 = 0 + 12 ANSWER x  6

Problem 3 Solution C2 + 2C - 8 = 0 Answer (C-2)(C+4) = 0 C  2 C-2 = 0 or C + 4 = 0 C-2+2 = 0 + 2 C + 4 - 4 = 0 - 4 C = 2 or C = -4 Answer C  2 C  -4