Algebra Know-It-ALL

the denominator. But there’s no good reason to expand the products of the binomials. That would produce

a formula with fewer grouping symbols, but the arithmetic would be more cumbersome in practical use.

Let’s express the formula as

x= {(b 2 −b 1 )± [(b 1 −b 2 )^2 − 4(a 1 −a 2 )(c 1 −c 2 )]1/2} / (2a 1 − 2 a 2 )

After we’ve found the x-values of the solutions by plugging in the coefficients, plugging in the constants, and

going through the arithmetic, we can put those x-values into either of the original two functions and calculate

they-values. If we call the x-values x 1 and x 2 , and if we use the first of the original functions, we have

y 1 =a 1 x 12 +b 1 x 1 +c 1

and

y 2 =a 1 x 22 +b 1 x 2 +c 1

The solutions of the whole system can be written as ordered pairs in the form (x 1 ,y 1 ) and (x 2 ,y 2 ).

Enter the Cubic

Let’s solve a two-by-two system in which one equation is linear and the other is cubic. Con-

sider these:

x^3 + 6 x^2 + 14 x−y=− 7

and

− 6 x+ 2 y= 2

First, we morph

Again, it appears as if we ought to let x be the independent variable, and then derive two func-

tions of that variable. In the first equation, we can add 7 to each side, getting

x^3 + 6 x^2 + 14 x−y+ 7 = 0

Then we can add y to each side and transpose the equation left-to-right, obtaining y as a

function of x:

y=x^3 + 6 x^2 + 14 x+ 7

In the second equation, we can divide through by −2 to obtain

3 x−y=− 1

Adding 1 to each side gives us

3 x−y+ 1 = 0

456 More Two-by-Two Systems