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Mastering Equation Solutions: Variables, Grouping Symbols & Identifying Equations

This resource highlights key concepts for solving equations with variables on each side, using grouping symbols, and recognizing cases of no solutions or identities. It provides step-by-step examples, including how to solve equations like 8 + 5s = 7s - 2 and 6 + 4q = 12q - 42. You'll learn to check your solutions effectively and understand when an equation has no solution or is an identity. Ideal for reinforcing classroom learning and preparing for practice tests, this guide helps build a solid foundation in algebraic problem-solving.

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Mastering Equation Solutions: Variables, Grouping Symbols & Identifying Equations

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

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

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

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

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

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

  7. A • B • C • D A. 38 B. 28 C. 10 D. 36 Lesson 5 CYP2

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

  9. B. Solve . No Solutions or Identity Original equation 4t + 80 = 4t + 80 Distributive 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. Lesson 5 Ex3

  10. A.true for all values of a B.no solution C. D.2 A. • A • B • C • D Lesson 5 CYP3

  11. A.true for all values of c B.no solution C. D.0 B. • A • B • C • D Lesson 5 CYP3

  12. Concept Summary 2-5

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

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

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

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

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

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

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