Lesson 5 Menu Five-Minute Check (over Lesson 2-4) Main Ideas and Vocabulary Targeted TEKS Example 1: Solve an Equation with Variables on Each Side Example.

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Lesson 5 Menu Five-Minute Check (over Lesson 2-4) Main Ideas and Vocabulary Targeted TEKS Example 1: Solve an Equation with Variables on Each Side Example 2: Solve an Equation with Grouping Symbols Example 3: No Solutions or Identity Concept Summary: Steps for Solving Equations Example 4: Test Example

Lesson 5 MI/Vocab identity Solve equations with the variable on each side. Solve equations involving grouping symbols.

Lesson 5 Ex1 Solve 8 + 5s = 7s – 2. Check your solution s = 7s – 2Original equation 8 + 5s – 7s = 7s – 2 – 7sSubtract 7s from each side. 8 – 2s = –2Simplify. 8 – 2s – 8 = –2 – 8Subtract 8 from each side. –2s = –10Simplify. Answer: s = 5Simplify. Solve an Equation with Variables on Each Side Divide each side by –2. To check your answer, substitute 5 for s in the original equation.

A.A B.B C.C D.D Lesson 5 CYP1 Solve 9f – 6 = 3f + 7. A. B. C. D.2 BrainPOP: Two Step Equations

Lesson 5 Ex2 Solve an Equation with Grouping Symbols 6 + 4q = 12q – 42Distributive Property 6 + 4q – 12q = 12q – 42 – 12qSubtract 12q from each side. 6 – 8q = –42Simplify. 6 – 8q – 6 = –42 – 6Subtract 6 from each side. –8q = –48Simplify. Original equation

Lesson 5 Ex2 Solve an Equation with Grouping Symbols Divide each side by –8. To check, substitute 6 for q in the original equation. Answer: q = 6Simplify.

Lesson 5 CYP2 1.A 2.B 3.C 4.D A.38 B.28 C.10 D.36

Lesson 5 Ex3 No Solutions or Identity A. Solve 8(5c – 2) = 10(32 + 4c). 8(5c – 2) = 10(32 + 4c)Original equation 40c – 16 = cDistributive Property 40c – 16 – 40c = c – 40cSubtract 40c from each side. –16 = 320This statement is false. Answer: Since –16 = 320 is a false statement, this equation has no solution.

Lesson 5 Ex3 No Solutions or Identity 4t + 80 = 4t + 80Distributive Property Answer: Since the expression on each side of the equation is the same, this equation is an identity. The statement 4t + 80 = 4t + 80 is true for all values of t. Original equation B. Solve.

1.A 2.B 3.C 4.D Lesson 5 CYP3 A. A.true for all values of a B.no solution C. D.2

1.A 2.B 3.C 4.D Lesson 5 CYP3 B. A.true for all values of c B.no solution C. D.0

Concept Summary 2-5

Lesson 5 Ex4 Find the value of H so that the figures have the same area. A 1B 3C 4D 5 Read the Test Item Solve the Test Item You can solve the equation or substitute each value into the equation and see if it makes the equation true. We will solve by substitution. represents this situation.

Lesson 5 Ex4 A: Substitute 1 for H. ? ?

Lesson 5 Ex4 B: Substitute 3 for H. ? ?

Lesson 5 Ex4 C: Substitute 1 for H. ? ?

Lesson 5 Ex4 D: Substitute 5 for H. Answer: Since the value 5 makes the statement true, the answer is D. ? ?

A.A B.B C.C D.D Lesson 5 CYP4 A.1 B.2 C.3 D.4 Find the value of x so that the figures have the same area.