If we call x the independent variable, then the quadratic equation is already expressed as a
function of x, so we don’t have to manipulate it. The linear equation can be rearranged so it appears
as a function of x by adding y to each side and then transposing the left and right sides, getting
y= 2 x+ 1
We now have two functions in which x is the independent variable and y is the dependent
variable.
Next, we mix
When we mix the independent-variable parts of the above functions, we have
x^2 +x− 5 = 2 x+ 1
We can subtract 1 from both sides to obtain
x^2 +x− 6 = 2 x
Then we can subtract 2x from both sides to get
x^2 −x− 6 = 0
Next, we solve
We’ve derived a quadratic equation that can be solved using any of the methods from
Chaps. 22 and 23. We’re lucky here, because this equation can be factored into
(x+ 2)(x− 3) = 0
The roots are found by solving the two first-degree equations
x+ 2 = 0
and
x− 3 = 0
giving us x=−2 or x= 3. We can substitute these two values for x into either of the morphed
original equations to obtain corresponding values for y. The simpler of the two is
y= 2 x+ 1
For x=−2, we have
y= 2 × (−2)+ 1
=− 4 + 1
=− 3
448 More Two-by-Two Systemss