Question

How do you solve `x - y = 4` and `y = x² - 34` using substitution?

Answer 1

Since one of the variables is already isolated in one of the equations (y; equation II), we can immediately substitute.

Explanation:

`x - (x^2 - 34) = 4`

`x - x^2 + 34 = 4`

`-x^2 + x + 30 = 0`

You can solve this by factoring. With trinomials `y = ax^2 + bx + c, a != 1`, you must find two numbers that multiply to `a xx c` and that add to b. Two numbers that multiply to -30 and that add to 1 are +6 and -5.

`-x^2 + 6x - 5x + 30 = 0`

`-x(x - 6) - 5(x - 6) = 0`

`(-x - 5)(x - 6) = 0`

`x = -5 and 6`

Substituting both solutions for x in the original equation we get:

`y = (-5)^2 - 34 or y = (6)^2 - 34`

`y = 25 - 34 or y = 36- 34`

`y = -9 or y = 2`

Thus, the solution set is `{-5, -9} and {6,2}`. Always make to put the right x value with the right y value and to place them in the brackets in the form `{x, y}`.

Practice exercises:

1. Solve the following systems by substitution. Leave answers in exact form when necessary.

a) `x^2 + 4 = y, 2x + 3y = -8`

b) `x^2 + 2y - 5 = 0, 3y - 5x = 6`

Good luck!