Question

Solving systems by substitution

Answer 1

9514 1404 393

Answer:

  (x, y) = (0, -2)

Step-by-step explanation:

When solving by substitution, you usually want to find an expression for one of the variables in terms of the other. So, the first thing you look for is an equation that has a coefficient of 1 or -1 on one of the variables. Recognizing that the second equation's terms all have a common factor of 3, you basically have two choices.

Substitute for y

Using equation 1, you can write an expression for y:

  y = 5x -2 . . . . . . add 5x to both sides

Then substituting this into the original equation 2, you have ...

  -3x +6(5x -2) = -12

  27x -12 = -12 . . . . . . . simplify

  27x = 0 . . . . . . . . . add 12

  x = 0 . . . . . . . . .  divide by 27

  y = 5(0) -2 = -2 . . . . find y using the expression for substitution

The solution is (x, y) = (0, -2).

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Substitute for x

If you decide you'd rather substitute for x, you can solve the second equation easily for x.

  -3x +6y = -12

  x -2y = 4 . . . . . . divide by -3

  x = 2y +4 . . . . . . add 2y

Substituting for x in the first equation gives ...

  -5(2y +4) +y = -2 . . . . substitute for x

  -9y -20 = -2 . . . . . . . simplify

  -9y = 18 . . . . . . . . .  add 20

  y = -2 . . . . . . . . . . . divide by -9

  x = 2(-2) +4 = 0 . . . . find x using the expression for substitution

The solution is (x, y) = (0, -2).

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Additional comment

In some cases, there are no variables that have a coefficient of ±1, so you just need to "bite the bullet" and deal with the resulting fractions.

Example:

  solve for y: -5x +2y = -2

     2y = 5x -2

     y = 5/2x -1 . . . . expression used to substitute for y

Of course, you can multiply the equation after substitution by 2 to eliminate fractions, or just work the problem as is. The point of looking for coefficients of ±1 is to avoid having to do arithmetic with fractions. It can help avoid errors to work with integers, but ultimately the method is the same regardless of the form of the numbers.

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You don't always have to substitute for the "bare" variable. Sometimes it can save steps to substitute for expressions instead of variables. If our system of equations were ...

• -5x +2y = -2
• -3x +6y = -12

You can substitute into the second equation for (2y). In that case, the second equation becomes ...

  -3x +3(2y) = -12

  -3x +3(5x-2) = -12 . . . . . . where 2y = 5x -2