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Module 8: Linear Equations and Linear Systems

Solving Using Substitution

The first method relies purely on isolations and plugging to solve. While we can’t determine the value of a variable right off the bat, we can figure out an equation that equals to a certain variable first, and then use that to solve for both variables eventually.

Tutorial 39: How to Use Substitution to Solve a Linear System

Example:

3 x + 4 y = 70 {"version":"1.1","math":"3x+4y=70"} and 3 x + 2 y = 50 {"version":"1.1","math":"3x+2y=50"}

 

  1. Pick one linear equation of the two and isolate for one variable (you may choose).

    3 x + 4 y = 70 {"version":"1.1","math":"3x+4y=70"}

    3 x = 70 − 4 y {"version":"1.1","math":"3x=70-4y"}

    x = 70 − 4 y 3 {"version":"1.1","math":"x=\dfrac{70-4y}{3}"}

  2. Substitute this equation into the other linear equation and solve for the other variable.

    3 x + 2 y = 50 {"version":"1.1","math":"3x+2y=50"}

    3 70 − 4 y 3 + 2 y = 50 {"version":"1.1","math":"3\dfrac{70-4y}{3}+2y=50"}

    70 − 4 y + 2 y = 50 {"version":"1.1","math":"70-4y+2y=50"}

    − 4 y + 2 y = 50 − 70 {"version":"1.1","math":"-4y+2y=50-70"}

    − 2 y = − 20 {"version":"1.1","math":"-2y= -20"}

    y = 10 {"version":"1.1","math":"y=10"}

  3. Substitute this new value into the equation obtained in step 1 to solve for the original variable.

    x = 70 − 4 y 3 {"version":"1.1","math":"x=\dfrac{70-4y}{3}"}

    x = 70 − 4 ( 10 ) 3 {"version":"1.1","math":"x=\dfrac{70-4(10)}{3}"}

    x = 70 − 40 3 {"version":"1.1","math":"x=\dfrac{70-40}{3}"}

    x = 30 3 {"version":"1.1","math":"x=\dfrac{30}{3}"}

    x = 10 → P O I : ( 10 , 10 ) {"version":"1.1","math":"x=10\rightarrow POI: (10,10)"}

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