alg1 equations packet solving linear equations

Alg1 Equations Packet

1

Solving Linear Equations

Golden Rule of Algebra:

“Do unto one side of the equal sign as you will do to the other…”

Whatever you do on one side of the equal sign, you MUST do the same exact thing on the other

side. If you multiply by -2 on the left side, you have to multiply by -2 on the other. If you subtract 15 from one

side, you must subtract 15 from the other. You can do whatever you want (to get the x by itself) as long as you

do it on both sides of the equal sign.

Solving Single Step Equations:

To solve single step equations, you do the opposite of whatever the operation is. The opposite of

addition is subtraction and the opposite of multiplication is division.

Solve for x:

1) x + 5 = 12 2) x – 11 = 19 3) 22 – x = 17

4) 5x = -30 5) (x/-5) = 3 6) ⅔ x = - 8

Solving Multi-Step Equations:

3x – 5 = 22 To get the x by itself, you will need to get rid of the 5 and the 3.

+5 +5 We do this by going in opposite order of PEMDAS. We get rid

of addition and subtraction first.

3x = 27 Then, we get rid of multiplication and division.

3 3

x = 9

We check the answer by putting it back in the original equation:

3x - 5 = 22, x = 9

3(9) - 5 = 22

27 - 5 = 22

22 = 22 (It checks)


2

Simple Equations:

1) 9x - 11 = -38 2) 160 = 7x + 6 3) 32 - 6x = 53

4) -4 = 42 - 4x 5) ¾x - 11 = 16 6) 37 = 25 - (2/3)x

7) 4x – 7 = -23 8) 12x + 9 = - 15 9) 21 – 4x = 45

10) (x/7) – 4 = 4 11) (-x/5) + 3 = 7 12) 26 = 60 – 2x

Equations with more than 1 x on the same side of the equal sign: You need to simplify (combine like terms) and then use the same steps as a multi-step equation.

Example:

9x + 11 – 5x + 10 = -15

9x – 5x = 4x and 4x + 21 = -15 Now it looks like a multistep eq. that we did in the 1st

11 + 10 = 21 -21 -21 Use subtraction to get rid of the addition.

4x = -36

4 4 Now divide to get rid of the multiplication

x = -9

13) 15x - 24 - 4x = -79 14) 102 = 69 - 7x + 3x 15) 3(2x - 5) - 4x = 33

16) 3(4x - 5) + 2(11 - 2x) = 43 17) 9(3x + 6) - 6(7x - 3) = 12

18) 7(4x - 5) - 4(6x + 5) = -91 19) 8(4x + 2) + 5(3x - 7) = 122

Equations with x's on BOTH sides of the equal sign: You need to "Get the X's on one side and the numbers on the other." Then you can solve.

Example: 12x – 11 = 7x + 9

-7x -7x Move the x’s to one side.

5x – 11 = 9 Now it looks like a multistep equation that we did in the 1st section.

+11 +11 Add to get rid of the subtraction.

5x = 20

5 5 Now divide to get rid of the multiplication

x = 4

20) 11x - 3 = 7x + 25 21) 22 - 4x = 12x + 126 23) ¾x - 12 = ½x -6

24) 5(2x + 4) = 4(3x + 7) 25) 12(3x + 4) = 6(7x + 2) 26) 3x - 25 = 11x - 5 + 2x


3

Solving Quadratic Equations

Solving quadratic equations (equations with x2 can be done in different ways. We will use two

different methods. What both methods have in common is that the equation has to be set to = 0. For instance, if

the equation was x2

– 22 = 9x, you would have to subtract 9x from both sides of the equal sign so the equation

would be x2 – 9x – 22 = 0.

Solve by factoring: After the equation is set equal to 0, you factor the trinomial.

x2 – 9x – 22 = 0

(x-11) (x+2) = 0

Now you would set each factor equal to zero and solve. Think about it, if the product of the two binomials

equals zero, well then one of the factors has to be zero.

x2 – 9x – 22 = 0

(x-11) (x+2) = 0

x – 11 = 0 x + 2 = 0

+11 +11 -2 -2

x = 11 or x = -2 * Check in the ORIGINAL equation!

Solving Quadratics by Factoring:

20) x2 - 5x - 14 = 0 21) x

2 + 11x = -30 22) x

2 - 45 = 4x

23) x2 = 15x - 56 24) 3x

2 + 9x = 54 25) x

3 = x

2 + 12x

26) 25x2 = 5x

3 + 30x 27) 108x = 12x

2 + 216 28) 3x

2 - 2x - 8 = 2x

2

29) 10x2 - 5x + 11 = 9x

2 + x + 83 30) 4x

2 + 3x - 12 = 6x

2 - 7x - 60


4

Solve using the quadratic formula:

When ax2 + bx + c = 0

x = -b ± √b2 – 4ac .

2a

a is the coefficient of x2

b is the coefficient of x

c is the number (third term)

Notice the ± is what will give your two answers (just like you had when solving by factoring)

x2 – 9x – 22 = 0 x = -b ± √b

2 – 4ac .

a = 1 2a

b= - 9

c = -22 x = -(-9) ± √ (-9)2 – 4(1)(-22) -4(1)(-22) = 88

2(1)

x = 9 ± √81 + 88

2

x= 9 ± √169 .

2

Split and do the + side and - side

9 + 13 9 – 13

2 2

x = 11 or x = -2

* Check in the ORIGINAL equation!

Solving Quadratics Using the Quadratic Formula:

31) 2x2 - 6x + 1 = 0 32) 3x

2 + 2x = 3 33) 4x

2 + 2 = -7x

34) 7x2 = 3x + 2 35) 3x

2 + 6 = 5x 36) 9x - 3 = 4x

2


5

Factor (Do not T it up… there is no equal sign so it is not an equation!!)

1) x2 + 4x + 4 2) x

2 – 6x + 9 3) x

2 - 18x + 81

4) x2 + 10x + 25 5) x

2 - 20x + 100 6) x

2 + 8x + 16

7) x2 – 22x + 121 8) x

2 + 32x + 256 9) x

2 – 40x + 400

Completing the Square Completing the square is another method that is used to solve quadratic equations. This method is especially helpful when the quadratic equation cannot be solved by simply factoring. ***Remember the standard form for a quadratic equation is: ax2 + bx + c = 0.***

Example: Steps: 1. – 1. Be sure that the coefficient of the highest exponent is 1. If it is not divide each term by that value to

create a leading coefficient of 1.

– 2. Move the constant term to the right hand side.

3. Prepare to add the needed value to create a perfect square trinomial. Be sure to balance the equation.

4. To create the perfect square trinomial:

a) Take

b) Add that value to both sides of the equation.

5. Factor the perfect square trinomial. 6. Rewrite the factors as a squared binomial. 7. Take the square root of both sides.

8. Split the solution into two equations 9. Solve for x. 10. Create your final answer.


6

More Examples: 1) 2) 3)

Example: Steps: 1. – 1. Be sure that the coefficient of the highest exponent is 1. If it is not divide each term by that value to

create a leading coefficient of 1.

– 2. Move the constant term to the right hand side.

3. Prepare to add the needed value to create a perfect square trinomial. Be sure to balance the equation.

4. To create the perfect square trinomial:

a) Take

b) Add that value to both sides of the equation.

5. Factor the perfect square trinomial. 6. Rewrite the factors as a squared binomial. 7. Take the square root of both sides.

8. Isolate X. Since you cannot combine it with

+5 +5

X = 5 9. Create your final answer


7

DO IN NOTEBOOK: 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18)

Alg1 Q4 Review ___: Solve each quadratic using completing the square:

1) 2)


8

3) 4)

5) 6)


9

7) 8)

9) 10)


10

11) 12)

13) 14)


11

15) 16)

17) 18)


12

19) 20)

21) x2 + 15x + 26 = 0 22) x

2 – 10x – 25 = 0


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Proportions and Percents

Proportions: A proportion is a statement that two ratios are equal. When trying to solve proportions we use the Cross

Products Property of Proportions.

A = C A(D) = B(C)

B D

Example:

6__ = x__ x + 5__ = 1.5___

11 121 12 6

6(121) = 11x 6(x + 5) = 12(1.5)

726 = 11x 6x + 30 = 18

-30 -30

726 = 11x 6x = -12

11 11 6 6

66 = x x = -2

1) x _ = 16 2) x – 3 _ = 12 _

14 35 x + 3 30

Percents: Is = %___

Of 100

Example:

What number is 20% of 50?

Is: ? x x = 20 .

Of: of 50 50 100

%: 20%

100: 100 100x = 20(50)

100x = 1,000

100x = 1,000

100 100

x = 10

a) What number is 40% of 160? b) 48 is what percent of 128?

c) 28 is 75% of what number? d) What number is 36% of 400?


14

Part I:

1) x = 18 . 2) 13 . = 65 . 3) x + 4 . = 6x .

12 54 x 90 9 18

4) - 16 . = 8 . 5) 14 . = 3x .

6x-2 11 16 3x + 3

6) What is 20% of 32? 7) 72 is 40% of what number?

8) 21.56 is what percent of 98? 9) - 31 is what percent of -124?

10) What is 62% of 140?

Part II: 1) x . = 13 . 2) - 13 . = 195 . 3) x + 4 . = 6x .

12 78 x 150 9 18

4) - 16 . = 8 . 5) x + 5 . = x . 6) x-4 _ = 9 _

5x-2 11 x - 3 9 12 x+8

7) 12 is 40% of what number?

8) 21.56 is what percent of 98? 9) 45 is what percent of 180?

10) What is 62% of 70?

Part III: 1) 23 . = 57.5 . 2) 3x – 5 . = 5x + 1 . 3) 5x -1 = 33 .

x 45 13 52 10x+5 45

4) x + 1 . = 2 . 5) 2x – 4 . = x - 2 . 6) x + 7 = x + 6 .

x + 6 x x + 5 x + 1 2x – 1 x - 2

10) What is 80% of 850? 8) 128 is 32% of what number?

9) 72 is what percent of 120? 10) What is 80% of 850?


15

Mixed Equations: Figure out what type of equation you have and then pick a strategy to solve.

1) 20 - (5/8)x = 40 2) 6(7x - 2) = 8(4x + 1) 3) 2(5x - 4) - 3(4x + 3) = -43

4) x2 + 44 = 15x 5) 3x

2 + 18x = 81 6) 3x

2 = 5x + 5

7) 11x - 5 = 7x - 53 8) 6(3x + 1) + 5(10 - 4x)= 39 9) ¼x - 33 = -49

10) 7x2 - 1 = 3x 11) 9(3x + 1) = 8(5x + 6) 12) 15x = x

2 – 16

13) x2 + 8x = 12 14) 9(4x + 7) - 6(7x + 10) = -54 15) 44 = 20 - 2x

16) 4x2 - 128 = 16x 17) 3x

2 - 8x + 6 = x + 6 18) 7(6x + 2) = 10(3x + 5)

19) 3x2 + 13x - 12 = 9x

2 - 11x - 12 20) 2x

2 - 14 = 10x

21) 14 . = 35 . 22) x + 5 . = x . 23) x - 10_ = 6 _

8x - 4 50 x - 4 32 12 x - 4

24) 10 . = 8 . 25) x - 6 . = x + 12 . 26) 2x - 3 = x - 3 _

7x + 2 5x + 4 2x - 3 x + 4 x + 1 x + 3


16

Systems of Equations Algebraically 1) 16x - 32y = 224

y = 2x – 1

2) 16x - 32y = 224

y = 2x – 1


17

Solving Systems of Equations Algebraically

In order to solve for two variables, you need to have two equations. If you only have one equation there are an

infinite amount of ordered pairs (x,y) that will work. For example:

4x – 2y = 16 you can have x = 4 and y = 0 (4,0) and (2, -2) and (0, -4) and an infinite amount of others. To be

able to solve for a single ordered pair, you need a second equation.

When we introduce the second equation, we will be able to solve for a single ordered pair that will work in

both equations. There are two ways to solve a system of equations (algebraically and graphically). We will

focus on solving algebraically. There are two methods of solving algebraically (substitution and elimination).

The key to both of them is changing one (or both) equations so there is only one variable to solve for. Then

you follow all the rules of solving for the one variable. Then plug the value back into one of the original

equations to find the value of the second variable. Always state your answer as an ordered pair.

SUBSTITUTION

Example: x = 3y + 8

5x + 2y = 6

Substitute 3y +8 for substitute the value Check in BOTH

the x in the 2nd equation for y back in to find x. ORIGINAL EQUATIONS!

5(3y + 8) + 2y = 6 y = -2 x = 3y + 8

Distribute and solve: x = 3(-2) + 8 (2) = 3(-2) + 8

15y + 40 + 2y = 6 x= -6 + 8 2 = -6 + 8

17y + 40 = 6 x = 2 2 = 2 (check)

17y = -34 5x + 2y = 6

17 17 (2, -2) State answer as an 5(2) + 2(-2) = 6

y = -2 ordered pair (x,y) 10 – 6 = 4

4 = 4 (check)

Solve each system and check (in both equations):

a) x = 2y + 1

5x – 6y = 13


18

b) y = 3x + 4

9x + 2y = -37

c) 4x + 2y = 24

10x + y = 8

d) 6x – 5y = 20

x + 3y = 11


19

e) 7x + 9y = -74

4x + y = -5

f) 8x + 3y = 35

10x – y = 1


20

Q4 Quiz ___ Review

1) y = 3x – 8

6x – 5y = 31

2) x = 2y + 9

5x + 8y = 117

3) y = 2x – 3

6x – 5y = 31


21

4) 4x + y = -9

12x – 7y = 123

5) x – 7y = 19

6x + 11y = 37

6) 3y – 2 = x

3x – 7y = -16

Answer Key: 1) (1,-5) 2) (17,4) 3) -4,-11) 4) (1.5, -15) 5) (12,-1) 6) (-17,-5)


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Solving Systems with Linear Combinations (“Elimination”):

Sometimes solving a system of equations using substitution can be very difficult. For these problems we solve

using Linear Combinations (or Elimination). With elimination you solve by eliminating one of the variables.

This is accomplished by adding the 2 equations together. Before you can add the equations together, you need

one of the two variables to have two things:

1) Same Coefficient

2) Different Signs (one positive and one negative)

When you add terms with the same coefficient and different signs, the term drops out. You then solve for the

variable that is left. After you have solved for one variable, you plug the value into one of the original

equations and solve for the 2nd

variable (just like Substitution). Then, you check the solution in both original

equations. The only difference between Substitution and Elimination is how you solve for the 1st variable.

After that they are the same.

Examples:

A) Sometimes it works out that the 2 equations already have a variable with the same coefficient and different

signs. You can then just add the equations:

3x + 4y = 10 (The +4y and -4y cancel out Plug x = -6 in: 3x+4y = 10

5x – 4y = -58 leaving you with just 8x.) 3(-6) + 4y = 10 3(-6) + 4(7) = 10

8x = -48 -18 + 4y = 10 -18 + 28 = 10

8 8 +18 +18 10 = 10(check)

4y = 28 5x – 4y = -58

x = -6 4 4 5(-6) – 4(7) = -58

y = 7 -30 – 28 = -58

-58 = -58(check)

Final Solution: (-6, 7) CHECK IN BOTH!!!!

B) Sometimes (usually) the equations do not have same coefficient and different signs, so we have a little bit of

manipulating to do.

3x + 8y = 25 With this system, nothing will drop out if we just add the

5x + 4y = 23 equations. So we will multiply the bottom one by (-2).

-2(5x + 4y = 23) Now the y’s have the same coefficient with different signs.

- 10x -8y = -46 3x + 8y = 25

3x + 8y = 25 Now plug x = 3 in: 3(3)+8(2) = 25

- 10x -8y = -46 3(3) + 8y = 25 9 + 16 = 25

- 7x = -21 9 + 8y = 25 25 = 25 (check)

-7 - 7 -9 -9 5x + 4y = 23

8y = 16 5(3) + 4(2) = 23

x = 3 8 8 15 + 8 = 23

y = 2 23 = 23 (check)

Final Solution: (3,2) CHECK IN BOTH!!!!


23

C. Sometimes we need to manipulate both equations. We can do this by

“criss crossing the coefficients.” 6x + 7y = 11 This is different than Example B, because no coeffcient

5x – 6y = -50 goes into another evenly.

You need the negative sign to change the 6x to negative -5(6x + 7y = 11) so the signs will be different.

6(5x – 6y = -50) You can also use 5 and -6.

You can also “criss cross” the y coefficients.

-30x – 35y = -55 Plug in y = 5 6x + 7y = 11

30x – 36y = -300 5x – 6(5) = -50 6(-4) + 7(5) = 11

- 71y = -355 5x – 30 = -50 -24 + 35 = 11

-71 -71 +30 +30 11 = 11 (check)

5x = -20 5x – 6y = -50

y = 5 5 5 5(-4) – 6(5) = -50

-20 – 30 = -50

x = -4 -50 = -50 (check)

Final Solution: (-4, 5) CHECK IN BOTH!!!!

Practice: 1) 7x + 3y = 10 2) 11x + 5y = 27 3) 9x + 7y = 126

5x – 6y = 56 4x + 6y = 60 7x – 9y = -32

4) 12x – 5y = 63 5) 5x + 9y = 14 6) 10x – 9y = 36

8x + 3y = 23 6x + 11y = 18 4x + 3y = -12

7) 5x + 6y = 42 8) 7x – 5y = -42 9) 4x – 3y = 19

3x + 14y = 20 8x + 3y = -48 8x + 5y = 159


24

Solve each system algebraically:

1) 5x - 2y = -9 2) -4x + 2y = -16

7x + 2y = -27 5x – 3y = 19

3) x = 2y -6 4) 5x – 6y = -74

5y –3x = 11 7x + 5y = 17

5) 4x – 5 = y 6) 7x + 4y = -11

7x + 5y = 83 5x + 2y = - 13

7) 5x – 6y = -17 8) x = 6 + 2y

3x + 8y = -16 6x – 5y = 15

9) 6x + 5y = 23 10) y = 3x + 4

11x + 4y = 6 8x – 9y = 59

11) 12x – 7y = 46 12) 9x – 4y = -88

4x + 3y = -6 2x + 5y = 4

13) 24x – 6y = -66 14) 5x – 6y = 42

12x – 3y = -33 15x – 18y = 54

15) 7x + 6y = -12 16) 13x – 3y = 78

5x + 2y = -20 4x + 6y = -66

17) 2y – 5 = x 18) 3x – 7y = -10

4x – 11y = -38 5x + 12y = -64

19) 6x – 17y = -104 20) 9x – 5y = -43

4x – 7y = -39 3x + 11y = 87

21) 9x = 11y + 25 22) 6y = 5x - 38

5x – 12y = 8 7x + 9y = 1

23) 6x + 5y = 33 24) y = 3x + 5

5x + 37 = 3y 12x – 7y = 1


25

Answer Key to Algebraic Systems (page 24):

1) (-3,-3) 7) (-4, -.5) 13) many sol. 19) (2.5, 7)

2) (5,2) 8) (0,-3) 14) no sol. 20) (-1/3, 8)

3) (8,7) 9) (-2,7) 15) (-6,5) 21) (4,1)

4) (-4,9) 10) (-5,-11) 16) (3,-13) 22) (4,-3)

5) (4,11) 11) (1.5, -4) 17) (7,6) 23) (-2,9)

6) (-5,6) 12)(-8, 4) 18) (-8, -2) 24) (-4,-7)

1) 6x – 5y = -7 2) 5x + 4y = -69

11x + 5y = 58 5x -7y = 52

3) 6x + 7y = -28 4) 11x – 4y = 53

5x – 14y = -182 7x – 8y = 1

5) 3x – 7y = 42 6) 9x – 4y = 177

2x + 5y = 57 6x – 5y = 111

7) 8x – 11y = 77 8) 13x – 2y = 72

6x + 4y = -28 9x + 5y = -14

9) 12x = 20- 8y 10) 5y = 8x + 97

5x – 6y = -57 10x + 7y = 51

Answer Key for this sheet:

1) (3, 5) 2) (-5, -11)

3) (-14, 8) 4) (7,6)

5) (21,3) 6) (21, 3)

7) (0, -7) 8) (4, -10)

9) (-3,7) 10) (-4, 13)


26

1) 7x – 4y = -86

9x – 4y = -98

2) 3x – 10y = -18

9x + 8y = -168

3) 5x + 8y = 70

-4x - 5y = -56


27

4) 10x + 11y = 37

8x – 7y = -160

5) 6x + 13y = -66

4x + 7y = -34

6) 5x - 9y = 22

8x – 5y = 101

Answer Key:

1) (-6,11) 2) (-16,-3) 3) (14,0) 4) (-9.5, 12) 5) (2,-6) 6) (17,7)


28

1) After beating the Mets with an RBI double, Bernie bought 4 slices of pizza and 2 cokes for $10.20.

After scoring on Bernie’s double, Matsui bought 3 slices of pizza and 3 cokes for $9.90. Find the price

of one coke. Find the price of 1 slice of pizza.

2) Mariano went to the donut shop and bought 6 donuts and 4 large coffees for $8.92. Tino went in right

after Mariano (congratulated him on Sunday's save) and bought 5 doughnuts and 6 large coffees for

$10.50. Find the price of 1 large coffee. Find the price of 1 donut. Robinson went in and bought 3

donuts and 2 large coffees. How much did he pay?

3) Derek and Jorge went to the burger stand and bought dinner. Derek had 2 cheeseburgers and 3 fries.

Jorge bought 3 cheeseburgers and 2 fries. Derek paid $16.55. Jorge paid $17.45. How much would 2

cheeseburgers and 1 fries cost?


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4) Alex and Tony went shopping for Mets clothing to set on fire. Alex bought 4 sweatshirts and 5 t-shirts

for $254. Tony bought 2 sweatshirts and 4 t-shirts for $154. How much would 2 sweatshirts and 3 t-

shirts cost?

5) Maggie and Erin went to see Frozen and went to the snack bar before finding their seats. Maggie paid

$11.05 for 2 candy bars and 3 sodas. Erin paid $17.55 for 3 candy bars and 5 sodas. Find the total cost

of 4 candy bars and 1 soda.

6) Sam and Peter went to the pizzeria and ordered some slices. Sam bought 2 slices of Sicilian and 2

regular and his bill was $10. Peter bought 3 slices of Sicilian and 1 regular for $10.50. How much

would 4 Sicilian and 5 slices of regular cost?


30

7) Solid ties cost $21 and striped ties cost $24. The store sold 200 ties and made $4,413. How many of

each were sold?

8) At a movie theater adult tickets cost $9.00 and child tickets cost $4.00. 120 people attended the last

showing of Silver Linings Playbook and $720 was collected at the ticket booth. How many of each

ticket was sold?

9) A jar of change was filled with only quarters and dimes. If there were 600 coins in the jar and there was

121.05 in the jar, how many of each coin were there?


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10) A 35-minute phone call cost $4.95. Introductory minutes cost $.16/min and additional minutes are

$.11/min. How many minutes were billed at each rate?

11) A 32-minute phone call cost $4.42. Introductory minutes cost $.17/min and additional minutes are

$.11/min. How many minutes were billed at each rate?

12) There was a jar of coins filled only with nickels and quarters. If there is $53.00 in the jar and there is a

total of 300 coins, how many of each coin are in the jar?

alg1 equations packet solving linear equations

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