Algebra Know-It-ALL

Are you confused?

We’ve found two ordered pairs (x,y) that solve the above pair of equations as a two-by-two system. How

do we know that these are the only two solutions for this system? We can demonstrate this visually for the

real-number solutions by graphing both equations together on the coordinate plane. The linear equation

shows up as a straight line, and the quadratic shows up as a parabola. In the next chapter, we’ll plot these

graphs. You’ll see that they intersect at the points corresponding to the ordered pairs (−2,−3) and (3, 7),

but nowhere else.

Here’s a challenge!

Suppose that a 1 ,a 2 ,b 1 ,b 2 , and c are real numbers, and neither a 1 nor a 2 are equal to 0. Consider these two

functions of x:

y=a 1 x+b 1

and

y=a 2 x^2 +b 2 x+c

Derive a general formula for solving this linear-quadratic system.

Solution

We can follow the same procedure as we did for the example we just solved, but we must use letter con-

stants instead of specific numbers, and we can’t simplify the expressions as much. These two equations are

ready to mix; we have no morphing to do. When we set the right sides equal to each other, we get

a 1 x+b 1 =a 2 x^2 +b 2 x+c

We can subtract b 1 from each side to get

a 1 x=a 2 x^2 +b 2 x+c−b 1

We can then subtract a 1 x from each side, obtaining

0 =a 2 x^2 +b 2 x+c−b 1 −a 1 x

The commutative, distributive, and grouping principles, followed by left-to-right transposition, allow us

to rearrange this equation to get

a 2 x^2 + (b 2 −a 1 )x+ (c−b 1 )= 0

If you’re confused by this rearrangement, or if you aren’t convinced that it’s correct, then check it out

for yourself. Change the subtractions to negative additions, move things around, and then change them

back again.

450 More Two-by-Two Systemss