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

1 Answer
Mar 29, 2016

Answer:

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!