ALGEBRA REVIEW PROBLEMS SOLVING LINEAR EQUATIONS HOW TO

ALGEBRA REVIEW PROBLEMS

SOLVING LINEAR EQUATIONS HOW TO SOLVE AN EQUATION 1. Distribute to remove the parenthesis. 2. Combine like terms on each side of the parenthesis. 3. Add/subtract to put all of the variables on one side of the equation. 4. Add/subtract to put all constants on the other side of the equation. 5. Multiply/divide to "undo" the coefficient with the variable. 6. You should now have your value for the variable.

1. 2(3 y + 4) = 32

1. 2(3 y + 4) = 32 6 y + 8 = 32 -8 -8 6 y = 24 6 6 y=4

2. 3(4 y – 5) = 69

2. 3(4 y – 5) = 69 12 y – 15 = 69 +15 12 y = 84 12 12 y=7

3. 36 = 3(7 x + 5)

3. 36 = 3(7 x + 5) 36 = 21 x + 15 -15 21 = 21 x 21 21 1=x

4. 156 = 4(6 w – 9)

4. 156 = 4(6 w – 9) 156 = 24 w – 36 +36 192 = 24 w 24 24 8=w

5. 6 x – (3 x + 7) = 29 6 x – 3 x – 7 = 29 +7 +7 3 x = 36 3 3 x = 12

6. 5 m – (6 m + 9) = 11 5 m – 6 m – 9 = 11 -m – 9 = 11 +9 +9 -m = 20 m = -20

7. 9(x + 8) + 2 = 8(x + 8) 9 x + 72 + 2 = 8 x + 64 9 x + 74 = 8 x + 64 -8 x x + 74 = 64 -74 x = -10

8. ¼(12 x + 36) – 4 = x + 27

8. ¼(12 x + 36) – 4 = x + 27 3 x + 9 – 4 = x + 27 3 x + 5 = x + 27 -x -x 2 x + 5 = 27 -5 -5 2 x = 22 2 2 x = 11

9. 5 x - 6 = ¼ (8 x + 36) 5 x – 6 = 2 x + 9 -2 x 3 x – 6 = 9 +6 +6 3 x = 15 3 3 x=5

10. 7(2 x – 3) = 3(4 x – 11)

10. 7(2 x – 3) = 3(4 x – 11) 14 x – 21 = 12 x – 33 -12 x 2 x – 21 = -33 +21 2 x = -12 2 2 x = -6

11. 5(x + 6) = 4(x + 7)

11. 5(x + 6) = 4(x + 7) 5 x + 30 = 4 x + 28 -4 x x + 30 = 28 -30 x = -2

12. 6(5 y + 7) = 7(4 y – 10)

12. 6(5 y + 7) = 7(4 y – 10) 30 y + 42 = 28 y – 70 -28 y 2 y + 42 = -70 -42 2 y = -112 2 2 y = -56

SOLVING SYSTEMS BY SUBSTITUTION Substitution method: easy to use when one of the equations is solved for one variable, such as x = or y =. a. b. c. d. Substitute all of the equation for x = or y = into the same variable in the other equation. Solve for that variable. Take that answer and substitute it into the first equation to find the value for that variable. Most of the time you will get answers for both variables. Occassionally both variables will be eliminated. If you end up with a true statement, such as 8 = 8, then you have coincident lines, thus there are infinitely many solutions. If you end up with a false statement, such as 8 = 12, then your lines are parallel and there is no solution.

1. y = 2 x x + y = 12

1. y = 2 x x + y = 12 Substitute 2 x for y in the second equation: x + 2 x = 12 Now solve for x: 3 x = 12 x=4 Now substitute x = 4 into the first equation: y = 2(4) y=8

2. x = 3 y – 1 x + 2 y = 9 Substitute 3 y – 1 for x in the 2 nd equation: 3 y – 1 + 2 y = 9 Now solve for y: 5 y – 1 = 9 5 y = 10 y=2 Now substitute y = 2 in the first equation: x = 3(2) – 1 x=5

3. y = 2 x – 5 4 x – y = 7 Substitute 2 x – 5 for y in the second equation: 4 x – (2 x – 5) = 7 4 x – 2 x + 5 = 7 2 x = 2 x=1 Now substitute x = 1 in the first equation: y = 2(1) – 5 y = -3

4. 2 x – 3 y = 12 x = 4 y + 1

4. 2 x – 3 y = 12 x = 4 y + 1 Substitute 4 y + 1 for x in the first equation: 2(4 y + 1) – 3 y = 12 8 y + 2 – 3 y = 12 5 y + 2 = 12 5 y = 10 y=2 Now substitute y = 2 in the second equation: x = 4(2) + 1 x=9

5. y = -x + 5 x – 4 y = 10 Substitute -x + 5 for y in the second equation: x – 4(-x + 5) = 10 x + 4 x – 20 = 10 5 x = 30 x=6 Substitute x = 6 for y in the first equation: y = -6 + 5 y = -1

6. x=y+2 4 x – 3 y = 11

6. x=y+2 4 x – 3 y = 11 Substitute y + 2 for x in the second equation: 4(y + 2) – 3 y = 11 4 y + 8 – 3 y = 11 y + 8 = 11 y=3 Now substitute y = 3 in the first equation: x=3+2 x=5

SOLVING SYSTEMS BY ELIMINATION Elimination method: easy to use when both equations are in standard form Ax + By = C. a. Look for both coefficients of the same variable to be opposites, such as 3 and -3. Add the two equations together, and one variable is eliminated. Solve for the other variable, then substitute to find the value for the previously eliminated variable. b. If you don't have both coefficients of the same variable as opposites, then you need to multiply one equation or both equations by a number or numbers to produce opposite coefficients. Then proceed as in substitution.

7. -2 x + 3 y = 14 x + 2 y = 7

7. -2 x + 3 y = 14 x + 2 y = 7 Notice that neither the x or the y will eliminate. Multiply the second equation by 2: -2 x + 3 y = 14 2 x + 4 y = 14 Now add the two equations together. The x is now eliminated: 7 y = 28 y=4 Now substitute y = 4 in either equation; I choose the second equation: x + 2(4) = 7 x+8=7 x = -1

8. 6 x – y = -4 2 x + 2 y = 15

8. 6 x – y = -4 2 x + 2 y = 15 Neither the x or the y will eliminate. Multiply the first equation by 2 since the signs are already opposite; or multiply the second equation by -3. I choose the 1 st option: 12 x – 2 y = -8 2 x + 2 y = 15 Now add the equations together; the y is eliminated. 14 x = 7 x = ½ or. 5 Substitute x = ½ into either equation; I choose the 2 nd: 2(½) + 2 y = 15 1 + 2 y = 15 2 y = 14 y=7

9. x+y=1 2 x – y = -2

9. x+y=1 2 x – y = -2 This time the y will eliminate; add the equations together: 3 x = -1 x=-⅓ Now substitute x in either equation; I choose the first: -⅓ + y = 1 y=1⅓

10. 5 x – 3 y = -11 4 x – 8 y = 8

10. 5 x – 3 y = -11 4 x – 8 y = 8 In this system, both equations need to be multiplied to be able to eliminate. You can choose to eliminate the x’s or the y’s. I choose the x’s. Multiply the first equation by 4 and the second equation by -5: 20 x – 12 y = -44 -20 x + 40 y = -40 Now add the equations together and the x is eliminated. 28 y = -84 y = -3 Substitute y = -3 in either equation; I choose the 2 nd equation: 4 x - 8(-3) = 8 4 x + 24 = 8 4 x = -16 x = -4

11. 5 x – 5 y = 15 6 x + 4 y = 13

11. 5 x – 5 y = 15 6 x + 4 y = 13 I choose to eliminate the y since the signs are already opposite. Multiply the first equation by 4 and the second equation by 5: 20 x – 20 y = 60 30 x + 20 y = 65 Add the two equations together and the y is eliminated: 50 x = 125 x = 2. 5 Now substitute x = 2. 5 in either equation. I choose the second equation: 6(2. 5) + 4 y = 13 15 + 4 y = 13 4 y = -2 y=-½

12. 8 x – 4 y = 64 -3 x + 6 y = -24

12. 8 x – 4 y = 64 -3 x + 6 y = -24 I choose to eliminate the y. Multiply first equation by 3 and second equation by 2: 24 x – 12 y = 192 -6 x + 12 y = -48 Add the two equations together and the y is eliminated: 18 x = 144 x=8 Substitute x = 8 in either equation. I choose the second equation: -3(8) + 6 y = -24 6 y = 0 y=0

MULTIPLYING BINOMIALS You should be able to use both the box method and FOIL, which is using the distributive property.

1. (2 x + 4)(x + 1) Using FOIL: 2 x 2 + 2 x + 4 2 x 2 + 6 x + 4

2. (3 x + 7)(x – 3) Using FOIL: 3 x 2 – 9 x + 7 x – 21 3 x 2 – 6 x – 21

3. (4 x – 2)(5 x – 1) Using FOIL: 20 x 2 – 4 x – 10 x + 2 20 x 2 – 14 x + 2

4. (2 x + 1)(9 x – 5) Using FOIL: 18 x 2 – 10 x + 9 x – 5 18 x 2 – x – 5

5. (7 x – 4)(3 x + 6) Using FOIL: 21 x 2 + 42 x – 12 x – 24 21 x 2 + 30 x – 24

6. (5 x – 8)(4 x – 4) Using FOIL: 20 x 2 – 20 x – 32 x + 32 20 x 2 – 52 x + 32

SOLVING A QUADRATIC EQUATION BY FACTORING HOW TO SOLVE A QUADRATIC EQUATION (has an x 2) 1. 2. 3. 4. 5. 6. Manipulate the equation until it equals zero. Factor out any greatest common factors. Factor the equation. Factoring is the opposite of FOIL. Take each factor (the part in the parenthesis) and set it equal to zero. Solve that equation. Your answers (also called roots or zeros) are where the equation, if graphed, would intersect the x-axis.

1. 2 x + 8 x + 12 = 0

1. x 2 + 8 x + 12 = 0 Factor it. The signs are both positive. (x + 2)(x + 6) = 0 Now set each parenthesis equal to 0. x+2=0 x+6=0 Now solve each equation for x: x = -2 x = -6 This means that the parabola will cross the x-axis at -2 and at -6. These are called the roots, or the zeros, or the xintercepts.

2. 2 x + 3 x = 10

2. x 2 + 3 x = 10 First you must manipulate the equation so that it equals zero. The 10 needs to move from the right side of the equation to the left side. Since it was +10, when you move it to the left side it becomes -10. x 2 + 3 x – 10 = 0 Factor it. The signs will be different because of the -10. (x + 5)(x – 2) = 0 Now set each parenthesis equal to 0. x+5=0 x– 2=0 Now solve each equation for x: x = -5 x=2 This means that the parabola will cross the x-axis at -5 and at 2. These are called the roots, or the zeros, or the x-intercepts.

3. 2 x = 14 x + 32

3. x 2 = 14 x + 32 First you must manipulate the equation so that it equals zero. The 14 x + 32 needs to move from the right side of the equation to the left side. Since it was +14 x + 32, when you move it to the left side it becomes -14 x – 32. X 2 – 14 x – 32 Factor it. The signs will be different because of the -32. (x + 2)(x – 16) = 0 Now set each parenthesis equal to 0. x+2=0 x – 16 = 0 Now solve each equation for x: x = -2 x = 16 This means that the parabola will cross the x-axis at -2 and at 16. These are called the roots, or the zeros, or the x-intercepts.

4. 2 x – 10 x + 9 = 0

4. x 2 – 10 x + 9 = 0 Factor it. The signs will be the same because of the +9, and will both be negative because of the -10 x. (x – 9)(x – 1) = 0 Now set each parenthesis equal to 0. x– 9=0 x– 1=0 Now solve each equation for x: x=9 x=1 This means that the parabola will cross the x-axis at 9 and at 1. These are called the roots, or the zeros, or the x-intercepts.

5. 2 2 x + 6 = 7 x

5. 2 x 2 + 6 = 7 x First you must manipulate the equation so that it equals zero. The 7 x needs to move from the right side of the equation to the left side. Since it was +7 x, when you move it to the left side it becomes -7 x. 2 x 2 – 7 x + 6 = 0 Factor it. The signs will be the same because of the +6, and both will be negative because of the -7 x. This one also has what we call a leading coefficient in front of the x 2. (2 x – 3)(x – 2) = 0 Now set each parenthesis equal to 0. 2 x – 3 = 0 x– 2=0 Now solve each equation for x: 2 x = 3 x = 1. 5 x=2 This means that the parabola will cross the x-axis at 1. 5 and at 2. These are called the roots, or the zeros, or the x-intercepts.

6. 2 x – 13 x + 36 = 0

6. x 2 – 13 x + 36 = 0 Factor it. The signs will be the same because of the + 36, and both will be negative because of the -13 x. (x – 9)(x – 4) = 0 Now set each parenthesis equal to 0. x– 9=0 x– 4=0 Now solve each equation for x: x=9 x=4 This means that the parabola will cross the x-axis at 9 and at 4. These are called the roots, or the zeros, or the x-intercepts.

7. 5 2 x = 9 x + 2

7. 5 x 2 = 9 x + 2 First you must manipulate the equation so that it equals zero. The 9 x needs to move from the right side of the equation to the left side, and the +2. Since it was +9 x, when you move it to the left side it becomes -9 x. The +2 on the right becomes a -2. 5 x 2 – 9 x – 2 = 0 Factor it. The signs will be the different because of the-2. This one also has what we call a leading coefficient in front of the x 2. (5 x + 1)(x – 2) = 0 Now set each parenthesis equal to 0: 5 x + 1 = 0 x– 2=0 Now solve each equation for x: 5 x = -1 x = -. 2 x=2 This means that the parabola will cross the x-axis at -. 2 and at 2. These are called the roots, or the zeros, or the x-intercepts.

8. 2 15 x + 7 x – 2 = 0

8. 15 x 2 + 7 x – 2 = 0 Factor it. The signs will be the different because of the -2. This one also has what we call a leading coefficient in front of the x 2. (5 x – 1)(3 x + 2) = 0 Now set each parenthesis equal to 0: 5 x – 1 = 0 3 x + 2 = 0 Now solve each equation for x: 5 x = 1 3 x = -2/3 This means that the parabola will cross the x-axis at. 2 and at -2/3. These are called the roots, or the zeros, or the x-intercepts.

9. 17 x + 5 = -6 2 x

9. 17 x + 5 = -6 x 2 First you must manipulate the equation so that it equals zero. The -6 x 2 needs to move from the right side of the equation to the left side. Since it was -6 x 2, when you move it to the left side it becomes 6 x 2 + 17 x + 5 = 0 Factor it. The signs will be the same because of the +5. This one also has what we call a leading coefficient in front of the x 2. (3 x + 1)(2 x + 5) = 0 Now set each parenthesis equal to 0: 3 x + 1 = 0 2 x + 5 = 0 Now solve each equation for x: 3 x = -1 2 x = -5 x=-⅓ x = 2. 5 This means that the parabola will cross the x-axis at - ⅓ and at 2. 5. These are called the roots, or the zeros, or the x-intercepts.

10. 3 9 x + 2 30 x = 24 x

10. 9 x 3 + 30 x 2 = 24 x First you must manipulate the equation so that it equals zero. The 24 x needs to move from the right side of the equation to the left side. Since it was +24 x, when you move it to the left side it becomes -24 x. 9 x 3 + 30 x 2 – 24 x = 0 First you need to look for a greatest common factor, the GCF. All terms in the trinomial have 3 x in common. Factor out the GCF. Then factor the remaining trinomial. The signs will be the different because of the -24 x. This one also has what we call a leading coefficient in front of the x 3. 3 x(3 x 2 + 10 x – 8) = 0 3 x(3 x – 2)(x + 4) = 0 Now set each factor and each parenthesis equal to 0: 3 x = 0 3 x – 2 = 0 x+4=0 Now solve each equation for x: x=0 3 x = 2 x = -4 x = 2/3 Since this has an x 3, this means that the parabola will cross the x-axis at 3 places: at 0, at 2/3 and at -4. These are called the roots, or the zeros, or the x-intercepts.

11. 3 8 x = 2 x

11. 8 x 3 = 2 x First you must manipulate the equation so that it equals zero. The 2 x needs to move from the right side of the equation to the left side. Since it was +2 x, when you move it to the left side it becomes -2 x. 8 x 3 – 2 x = 0 First you need to look for a greatest common factor, the GCF. All terms in the trinomial have 2 x in common. Factor out the GCF. Then factor the remaining binomial. The signs will be the different because of the -2 x. This one also has what we call a leading coefficient in front of the x 3. 2 x(4 x 2– 1) = 0 This binomial is called a “difference of two squares”. 2 x(2 x – 1)(2 x + 1) = 0 Now set each factor and each parenthesis equal to 0: 2 x = 0 2 x – 1 = 0 2 x + 1 = 0 Now solve each equation for x: x=0 2 x = 1 2 x = -1 x=½ x=-½ Since this has an x 3, this means that the parabola will cross the x-axis at 3 places: at 0, at ½ and at - ½. These are called the roots, or the zeros, or the x-intercepts.

12. 3 5 x = 2 40 x – 80 x

12. 3 5 x = 2 40 x – 80 x First you must manipulate the equation so that it equals zero. The 40 x 2 needs to move from the right side of the equation to the left side and it becomes a -40 x 2. The 80 x needs to move from the right side of the equation to the left side. Since it was – 80 x, when you move it to the left side it becomes 80 x. 5 x 3 – 40 x 2 + 80 x = 0 First you need to look for a greatest common factor, the GCF. All terms in the trinomial have 5 x in common. Factor out the GCF. Then factor the remaining trinomial. The signs will be the same because of the +80 x, and both signs will be negative because of the – 40 x 2. This one also has what we call a leading coefficient in front of the x 3. 5 x(x 2 + 8 x + 16) = 0 5 x(x – 4)2 Now set each factor and each parenthesis equal to 0: 5 x = 0 x– 4=0 Now solve each equation for x: x=0 x=4

13. 3 x – 16 x = 0

13. x 3 – 16 x = 0 First you need to look for a greatest common factor, the GCF. All terms in the trinomial have x in common. Factor out the GCF. Then factor the remaining binomial. The signs will be the different because of the -16 x. x(x 2 – 16) = 0 This binomial is called a “difference of two squares”. x(x – 4)(x + 4) = 0 Now set each factor and each parenthesis equal to 0: x=0 x– 4=0 x+4=0 Now solve each equation for x: x=0 x=4 x = -4 Since this has an x 3, this means that the parabola will cross the xaxis at 3 places: at 0, at 4 and at - 4. These are called the roots, or the zeros, or the x-intercepts.