How do you solve #x-2y=2# and #y^2-x^2=2x+4#?

1 Answer
Mar 24, 2015

The logic behind solving by substitution is expressing the given equation in two variables (say x and y) in terms of only one variable (i.e. make only x or y appear as the variable in the equation).

Usually, you will be given two equations that have been expressed in terms of two variables. Do not panic!

  1. Rearrange one of them to express one of the variables in terms of the other. Usually, it is easier to select a linear equation for this purpose.
    In this case, write #x-2y=2# as #x=2+2y#. Note that we had to add 2y on both sides of the equation to get that result.
  2. Substitute the value of x into the other equation so that you get an expression in terms of y.
    So #y^2-x^2=2x+4# becomes #y^2-(2+2y)^2=2(2+2y)+4#
    #=># #y^2 - (4+8y+4y^2)=4+4y+4#
    #=># #-3y^2 -4-8y=8+4y#
    #=># #-3y^2 -12-12y=0#
    #=># #y^2+4y+4=0# (Taking -3 as the common factor)
    #=># #(y+2)^2=0#
    #=># #y=-2#
  3. Resubstitute the value of y into the original linear equation to get the value of x.
    So #x=2-4# #=># #x=-2#