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**Identify the number of solutions of an equation**

EXAMPLE 4 Identify the number of solutions of an equation Solve the equation, if possible. a. 3x = 3(x + 4) b. 2x + 10 = 2(x + 5) SOLUTION a. 3x = 3(x + 4) Original equation 3x = 3x + 12 Distributive property The equation 3x = 3x + 12 is not true because the number 3x cannot be equal to 12 more than itself. So, the equation has no solution. This can be demonstrated by continuing to solve the equation.

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**Identify the number of solutions of an equation**

EXAMPLE 4 Identify the number of solutions of an equation 3x – 3x = 3x + 12 – 3x Subtract 3x from each side. 0 = 12 Simplify. ANSWER The statement 0 = 12 is not true, so the equation has no solution.

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**Identify the number of solutions of an equation**

EXAMPLE 1 EXAMPLE 4 Identify the number of solutions of an equation b. 2x + 10 = 2(x + 5) Original equation 2x + 10 = 2x + 10 Distributive property ANSWER Notice that the statement 2x + 10 = 2x + 10 is true for all values of x.So, the equation is an identity, and the solution is all real numbers.

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**GUIDED PRACTICE for Example 4 8. 9z + 12 = 9(z + 3) SOLUTION**

Original equation 9z + 12 = 9z + 27 Distributive property The equation 9z + 12 = 9z + 27 is not true because the number 9z + 12 cannot be equal to 27 more than itself. So, the equation has no solution. This can be demonstrated by continuing to solve the equation.

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**The statement 12 = 27 is not true, so the equation has no solution.**

GUIDED PRACTICE for Example 4 9z – 9z + 12 = 9z – 9z + 27 Subtract 9z from each side. 12 = 27 Simplify. ANSWER The statement 12 = 27 is not true, so the equation has no solution.

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**GUIDED PRACTICE for Example 4 9. 7w + 1 = 8w + 1 SOLUTION – w + 1 = 1**

Subtract 8w from each side. – w = 0 Subtract 1 from each side. ANSWER w = 0

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**GUIDED PRACTICE for Example 4 10. 3(2a + 2) = 2(3a + 3) SOLUTION**

Original equation 6a + 6 = 6a + 6 Distributive property ANSWER The statement 6a + 6 = 6a + 6 is true for all values of a. So, the equation is an identity, and the solution is all real numbers.

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EXAMPLE 2 Solving an Equation Involving Decimals 1.4x –1.8 + 2.35x = 0.21 Original equation. (1.4x –1.8 + 2.35x)100 = (0.21)100 Multiply each side by 100.

EXAMPLE 2 Solving an Equation Involving Decimals 1.4x –1.8 + 2.35x = 0.21 Original equation. (1.4x –1.8 + 2.35x)100 = (0.21)100 Multiply each side by 100.

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