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EXAMPLE 2 Checking Solutions Tell whether (7, 6) is a solution of x + 3y = 14. – x + 3y = 14 Write original equation. 7 + 3( 6) = 14 – ? Substitute 7 for x and 6 for y. – 7 + ( 18) = 14 – ? Simplify. 11 = 14 – Solution does not check. ANSWER The ordered pair (7, 6) is not a solution of x + 3y = 14. –

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EXAMPLE 3 Finding Solutions of an Equation Write the equation 4x + y = 15 in function form. Then list four solutions. SOLUTION STEP 1 Rewrite the equation in function form. 4x + y = 15 Write original equation. y = 15 4x– Subtract 4x from each side.

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EXAMPLE 3 Finding Solutions of an Equation STEP 2 Choose several values to substitute for x. Then solve for y. x -value Substitute for x. Evaluate.Solution x = 0 x = 1 x = 2 y = 19 y = 15 y = 11 y = 7 (0, 15) (1, 11) (2, 7) x = 1 – ( 1, 19) – y = 15 4( 1) –– y = 15 4(0) – y = 15 4(2) – ANSWER Four solutions are ( 1, 19), (0, 15), (1, 11), and (2, 7). – – y = 15 4(1)

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GUIDED PRACTICE for Examples 2 and 3 Tell whether the ordered pair is a solution of the equation. ANSWER The ordered pair ( 6, 5 ) is not a solution of y = 3x 7. – 2. y = 3x 7; (6, 5) –

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GUIDED PRACTICE for Examples 2 and 3 ANSWER The ordered pair ( 4, 1 ) is a solution of –2x – 4y = 12. –– 2x 4y = 12; (–4, 1) –– – 3.

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GUIDED PRACTICE for Examples 2 and 3 List four solutions of the equation. y = 2x + 6– 4. x -value Substitute for x. Evaluate.Solution x = 0 x = 1 y = 10 y = 8 y = 6 y = 4 (0, 6) (1, 4) x = 2 – x = 1 – ( 2, 10) – ( 1, 8) – y = 2 2 + 6 – – – y = 2 1 + 6 – y = 2 0 + 6 – y = 2 1 + 6 – ANSWER

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GUIDED PRACTICE for Examples 2 and 3 5. 3x + y = 4 x -value Substitute for x. Evaluate.Solution x = 0 x = 1 y = 10 y = 7 y = 4 y = 1 (0, 4) (1, 1) x = 2 – x = 1 – y = 4 3 0 – y = 4 3 1 – y = 4 3 2 –– y = 4 3 1 –– ( 2, 10) – ( 1, 7) – ANSWER

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