Algebra I Unit 1: Solving Equations in One Variable.

Algebra I Unit 1: Solving Equations in One Variable

Unit 1: Learning Goal 1.4

• Students will solve one-, two- and multi-step equations using the properties of real numbers and the properties of equality to justify the steps including real-world situations; represent solution process using concrete models.

Unit 1: Learning Goal 1.4 – Essential Knowledge and Skills for

Students• The students will use problem solving, reasoning,

communication, connections and representation to:– Using the properties of real numbers and the properties of

equality, and solve equations.– Use concrete and pictorial models to represent and solve

equations– Justify the steps to the solutions using the properties of real

numbers and the properties of equality.– Demonstrate an understanding that equivalent equations have

the same solution.– Demonstrate and understanding that adding the same

quantity to each side of an equation produces an equivalent equation.

– Write equations that represent real-world applications and solve using appropriate methods.

– Math/write a verbal statement that describes a given equation.

Unit 1: Learning Goal 1.4Algebra I Standards: EA-4.7, EA-1.3, and EA-

1.6

• EA-4.7 – Carry out procedures to solve linear equations in one variable algebraically.

• EA-1.3 – Apply algebraic methods to solve problems in real world contexts.

• EA-1.6 – Understand how algebraic relationships can be represented in concrete models, pictorial models, and diagrams.

Solving Equations by using addition and subtraction


27 -0.8 8Note: that when you have a term such as – c = -27 such as the

one above. The variable actually has a coefficient of -1. Remember that the product of two negative numbers is positive

you can multiply each side of the equation by -1 so that your answer is c = 27.


Helpful Hints: •If the same number is added or subtracted to each side of an equation, then the result is an equivalent equation. Equivalent Equations have the same solution.

•To Solve an equation means to find all values of the variable that make the equation a true statement. One way to do this is to isolate the variable having a coefficient of 1 on one side of the equation. You can sometimes do this by using the Addition Property of Equality.

•Isolating Variables – when isolating variables it does not matter whether the variable ends up on the right side or the left side of the equation. For example, the solution of 8 = 15 + z is still -7, even though the final step may be -7 = z.


Click here to try and electronically balance equations

http://nlvm.usu.edu/en/NAV/frames_asid_324_g_3_t_2.html?open=instructions&from=topic_t_2.html

http://nlvm.usu.edu/en/NAV/frames_asid_324_g_3_t_2.html?open=instructions&from=topic_t_2.html

Solving Equations by using multiplication and division


6y/6 = 54/6y = 9

-7d/ -7 = -84/-7d = 12

22b/22 = 176/22b = 8

2.4f/2.4 = 21.6/2.4f = 9


(-6)(p/-6) = (7/12)(-6)p = -3 1/2

Change – 2 1/3 into – 7/3, which

is an improper fraction (-3/7)(-7/3q) = (21)(-3/7)

q = -9

Solving Multi-Step Equations


10 – 7p = -18-10 -10 -7p = -28 -7p / -7 = -28 / -7 p = 4

- 1.9r + 9.3 = 15 - 9.3 = -9.3 -1.9r = 5.7 -1.9r / -1.9 = 5.7/-1.9 r = -3


(4)- 6 = [(-2n-3)/4](4)-24 = -2n-3-3 -3

-21/ -2 = - 2n/-2n = 10 1/2

(5)(t / 5 – 4)= - 10(5) t – 20 = -50

+ 20 = + 20 t = -30

Solving Equationswith variable on each side


Steps for solving Equations:Steps for solving Equations:Step 1: Use the Distributive Property to remove the grouping symbols.Step 2: Simplify the expressions on each side of the equals sign.Step 3: Use the Addition and/or Subtraction Properties of Equality to get the variables on one side of the equals sign and the numbers without variables on the other side of the equals sign.Step 4: Simplify the expression on each side of the equals sign.Step 5: Use the Multiplication or Division Property of Equality to solve.

•If the solution results in a false statement, there is no solution of the equation.•If the solution results in an identity, the solution is all numbers.


18 = 2n - 9

27 / 2 = 2n /2n = 13 1/2

+ 9 = + 9

18 + 2n = 4n – 9- 2n = - 2n

1 – 2.7 y = y

1/3.7 = 3.7y/3.7y = 1/3.7 or .27

+ 2.7y = + 2.7y

10 - 2.7y = y + 9- 9 = - 9


19.4 c – 2.4 = 6.4

19.4 c/19.4 = 8.8/19.4c = 8.8/19.4 or 0.45

+ 2.4 = + 2.4

11.1c – 2.4 = -8.3c + 6.4+ 8.3 c = + 8.3 c

3 = 12x + 8

- 5/12 = 12x/12x = -5/12

- 8 = - 8

3 – 4x = 8x + 8+ 4x = 4x

Work Cited

• Carter, John A., et. al. Glencoe Mathematics Algebra I. New York: Glencoe/McGraw-Hill, 2003.

• Greenville County Schools Math Curriculum Guide

Algebra I Unit 1: Solving Equations in One Variable.

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Transcript of Algebra I Unit 1: Solving Equations in One Variable.