Example 1 Solve an Equation with Variables on Each Side Solve 8 + 5c = 7c – 2. Check your solution. 8 + 5c = 7c – 2Original equation Answer: c = 5Simplify.

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Example 1 Solve an Equation with Variables on Each Side Solve 8 + 5c = 7c – 2. Check your solution c = 7c – 2Original equation Answer: c = 5Simplify. Divide each side by –2. To check your answer, substitute 5 for c in the original equation. – 7c = – 7cSubtract 7c from each side. 8 – 2c = –2Simplify. – 8 = – 8Subtract 8 from each side. –2c = –10Simplify.

A.A B.B C.C D.D Example 1 Solve 9f – 6 = 3f + 7. A. B. C. D.2

Example 2 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

Example 2 Solve an Equation with Grouping Symbols Divide each side by –8. To check, substitute 6 for q in the original equation. Answer: q = 6 q = 6Original equation

A.A B.B C.C D.D Example 2 A.38 B.28 C.10 D.36

Example 3 Find Special Solutions 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.

Example 3 Find Special Solutions 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.

A.A B.B C.C D.D Example 3 A. B.2 C.true for all values of a D.no solution

A.A B.B C.C D.D Example 3 B. A. B.0 C.true for all values of c D.no solution

Concept

Example 4 Find the value of H so that the figures have the same area. A 1B 3C 4D 5 Read the Test Item represents this situation.

Example 4 Solve the Test Item You can solve the equation or substitute each value into the equation and see if it makes the equation true.. 3H = 5(H – 2) 3H = 5H H + 10 = 5H -3H 10 = 2H = H

Example 4 Check: Substitute 5 for H. ? ?

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