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

Day 1

Solve an Inequality w < 3 All numbers less than 3 are solutions to this problem!

More Examples r -10 All numbers greater than-10 (including -10) ≥

More Examples 2 2 x > -1 All numbers greater than -1 make this problem true!

More Examples h ≤ h ≤ 8 All numbers less than 8 (including 8)

Solve the inequality on your own. 1.x + 3 > d > x - 8 < c – 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 the inequality and graph the solutions. y ≤ 4y + 18 –y 0 ≤ 3y + 18 –18 – 18 –18 ≤ 3y To collect the variable terms on one side, subtract y from both sides. Since 18 is added to 3y, subtract 18 from both sides to undo the addition. Since y is multiplied by 3, divide both sides by 3 to undo the multiplication. y  –6

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 m < 9 Since m is multiplied by 2, divide both sides by 2 to undo the multiplication. Solve the inequality and graph the solutions.

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

Solve the inequality and graph the solutions. Check your answer. 5t + 1 < –2t – 6 +2t +2t 7t + 1 < –6 – 1 < –1 7t < –7 7 t < –1 –5 –4 –3–2 – 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. Since t is multiplied by 7, divide both sides by 7 to undo the multiplication. The solution set is {t:t < –1}.

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

Day 2

Solving Compound Inequalities

Multiple Choice Solve:

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 –11 3x x -6 x -3

Solve “Tree it up”: 5w + 3 = 7 OR 5w + 3 = -7 Solve both equations for w 5w + 3 = 75w + 3 = -7 5w = 45w = -10 w = w = -2

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

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

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

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

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

Examples 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. x – 4 is positive x – 4  5 x – 4 is negative x  9 Case 1: Case 2: x – 4  –5 x  –1 solution: –1  x  9

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

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

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

Day 3

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

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

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

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

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

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

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. Like variables must be lined under each other.

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

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

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 1 st equation, the x’s would be eliminated when we add. So we will multiply the 1st equation by a – 2. Like variables must be lined under each other.

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

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

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

Laws of Exponents

Zero Exponents A nonzero based raise to a zero exponent is equal to one a 0 = 1

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

Basic Examples

Examples

Example Write 7,200,000 in scientific notation Big Number means Positive Exponent 7.2  10 6

Example Write 476 in scientific notation. Big Number means Positive Exponent 4.76  10 2

Example Write 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 Saturn is about 875,000,000 miles from the sun. What is this distance in scientific notation? ,600,  10 8

Day 4

= ±2 = ±4 = ±5 = ±10 = ±12

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

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

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

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

Solve using square roots. x 2 = –49 There is no real number whose square is negative. Answer: There is no real solution.

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

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

x 2 = –16 There is no real number whose square is negative. There is no real solution. Solve using square roots. Check your answer.

Solve using square roots. x = 7 –7 x = 7 x 2 = 0 The solution is 0. Subtract 7 from both sides. Take the square root of both sides.

Solve using square roots. 16x 2 – 49 = Add 49 to both sides. 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

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

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

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

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 4. Subtract the following polynomials: (9y - 7x + 15a) - (-3y + 8x - 8a)

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

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

Find the sum or difference. (5a – 3b) + (2a + 6b) 1.3a – 9b 2.3a + 3b 3.7a + 3b 4.7a – 3b

Find the sum or difference. (5a – 3b) – (2a + 6b) 1.3a – 9b 2.3a + 3b 3.7a + 3b 4.7a – 9b

Multiply (y + 4)(y – 3) 1.y 2 + y – 12 2.y 2 – y – 12 3.y 2 + 7y – 12 4.y 2 – 7y – 12 5.y 2 + y y 2 – y y 2 + 7y y 2 – 7y + 12

Multiply (2a – 3b)(2a + 4b) 1.4a ab – 12b 2 2.4a 2 – 14ab – 12b 2 3.4a 2 + 8ab – 6ba – 12b 2 4.4a 2 + 2ab – 12b 2 5.4a 2 – 2ab – 12b 2

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

Multiply (2p + 1)(p 2 – 3p + 4) 1.2p 3 + 2p 3 + p y 2 – y – 12 3.y 2 + 7y – 12 4.y 2 – 7y – 12

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

Multiply each: 4a 4 + 2a 3 + a - 1 Distribute the binomial 4a 4 + 2a 3 – 2a 2 + 2a 2 + a – 1

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

Day 5

Review: What is the GCF of 25a 2 and 15a? 5a Let’s go one step further… 1) FACTOR 25a a. Find the GCF and divide each term 25a a = 5a( ___ + ___ ) Check your answer by distributing. 5a3

2) Factor 18x x 3. Find the GCF 6x 2 Divide each term by the GCF 18x x 3 = 6x 2 ( ___ - ___ ) Check your answer by distributing. 32x

3) Factor 28a 2 b + 56abc 2. GCF = 28ab Divide each term by the GCF 28a 2 b + 56abc 2 = 28ab ( ___ + ___ ) Check your answer by distributing. 28ab(a + 2c 2 ) a2c 2

Factor 20x xy 1.x(20 – 24y) 2.2x(10x – 12y) 3.4(5x 2 – 6xy) 4.4x(5x – 6y)

5) Factor 28a b - 35b 2 c 2 GCF = 7 Divide each term by the GCF 28a b - 35b 2 c 2 = 7 ( ___ + ___ - _______ ) Check your answer by distributing. 7(4a 2 + 3b – 5b 2 c 2 ) 4a 2 5b 2 c 2 3b

Factor 16xy y 2 z + 40y 2 1.2y 2 (8x – 12z + 20) 2.4y 2 (4x – 6z + 10) 3.8y 2 (2x - 3z + 5) 4.8xy 2 z(2 – 3 + 5)

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)t 2 – 4t – 212) x x )x 2 –10x + 244) x 2 + 3x – 18 5) 2x 2 + x – 216) 3x x + 10 Factor These Trinomials! (t – 7)(t + 3)(x + 8)(x + 4) (x – 6)(x - 4)(x + 6)(x - 3) 2x 2 + 7x – 6x – 21 x(2x + 7) – 3(2x + 7) (x –3)(2x + 7) 3x 2 + 6x + 5x x(x + 2) + 5(x + 2) (x + 2)(3x + 2) REMEMBER YOU CAN CHECK YOUR ANSWER BY FOILING BACK OUT!!!!! , , , ,

Solve this proportion. ____ = ____ x Now cross multiply 5x = x = 36

Solve the proportion. ____ = ____ 3 2 x 18 Now cross multiply 2x = x =27

Just solve….. 10) 6x = x = 24

11) 16m = 8 16 m= Now Reduce

Use cross multiplying to solve the proportion = 45 t 2. x9x9 = = r n 10 = 28 8 t = 36 x = 3 r = 24 n = 35

Simplify: Factor the numerator and denominator Divide out the common factors. Write in simplified form.

Simplify: Factor the numerator and denominator Divide out the common factors. Write in simplified form.

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

Simplify: Factor the numerator and denominator Divide out the common factors. Write in simplified form.

Simplify: Factor the numerator and denominator Divide out the common factors. Write in simplified form.

Simplify each: