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Equation Solutions Checker

Check if ordered pairs are solutions to given equations. Find solutions by substituting values and evaluating the equation. Support for multiple equations.

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Equation Solutions Checker

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

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

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

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

  5. – – – 3. 2x 4y = 12; (–4, 1) ANSWER – – The ordered pair( 4 , 1)is a solution of–2x –4y= 12. for Examples 2 and 3 GUIDED PRACTICE

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

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

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