Algebra Know-It-ALL

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