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This module introduces the concept of substitution to solve simultaneous equations.

Here is the algorithm for substitution.

  • Solve one of the equations for one variable.
  • Plug this variable into the other equation.
  • Solve the second equation, which now has only one variable.
  • Finally, use the equation you found in step (1) to find the other variable.

Solving simultaneous equations by substitution

3x + 4y = 1 size 12{3x+4y=1} {}

2x y = 8 size 12{2x - y=8} {}

  • The easiest variable to solve for here is the y size 12{y} {} in the second equation.
    • y = 2x + 8 size 12{ - y= - 2x+8} {}
    • y = 2x 8 size 12{y=2x - 8} {}
  • Now, we plug that into the other equation:
    • 3x + 4 ( 2x 8 ) = 1 size 12{3x+4 \( 2x - 8 \) =1} {}
  • We now have an equation with only x size 12{y} {} in it, so we can solve for x size 12{y} {} .
    • 3x + 8x 32 = 1 size 12{3x+8x - "32"=1} {}
    • 11 x = 33 size 12{"11"x="33"} {}
    • x = 3 size 12{x=3} {}
  • Finally, we take the equation from step (1), y = 2x 8 size 12{y=2x - 8} {} , and use it to find y size 12{y} {} .
    • y = 2 ( 3 ) 8 = 2 size 12{y=2 \( 3 \) - 8= - 2} {}

So ( 3, 2 ) size 12{ \( 3, - 2 \) } {} is the solution. You can confirm this by plugging this pair into both of the original equations.

Why does substitution work?

We found in the first step that y = 2x 8 size 12{y=2x - 8} {} . This means that y size 12{y} {} and 2x 8 size 12{2x - 8} {} are equal in the sense that we discussed in the first chapter on functions—they will always be the same number, in these equations—they are the same . This gives us permission to simply replace one with the other, which is what we do in the second (“substitution”) step.

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Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
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