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Solving Algebraically by Substitution

Learn to solve algebraic equations by substitution or elimination methods step by step. Identify isolated variables, plug into equations, and simplify for solutions. Practice with examples provided to master the process.

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Solving Algebraically by Substitution

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  1. Solving Algebraically by Substitution Look for an isolated variable or one that is easy to isolate. Then plug in to other equation. y = -½x – 7 2x – 6y = 12 2x – 6 (-½x – 7) = 12 2x + 3x + 42 = 12 y = -½x – 7 5x = - 30 y = -½(-6) – 7 x = - 6 y = 3 – 7 (-6, - 4) y = - 4

  2. Solving Algebraically by Elimination I sure do love that trusty substitution! Look for a pair of matching opposites x + 3y = 8 6x – 3y = 27 + 7x = 35 7 7 But sometimes there’s an easier way! x + 3y = 8 x = 5 5 + 3y = 8 (5, 1) 3y = 3 y = 1

  3. Anyway, where were we? If you don’t have matching opposites, you can make them match. 4( ) 2x + y = 3 7x – 4y = 18 8x + 4y = 12 7x – 4y = 18 Find an LCM to distribute

  4. Steps to Solving by Elimination Look for equal opposites or a pair that’s easy to manipulate Step 1 Distribute the coefficient you need across one or both equations to make equal opposites Step2 Add equations together and cancel pair Step3 Step4 Solve for one variable Plug answer back in to find other variable Step5

  5. Practice 3x + 4y = 1 2x + 4y = 6 3x + 2y = -1 4x – 5y = -32 (-5, 4) (-3, 4)

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