Algebra Know-It-ALL

526 Review Questions and Answers

When we plug either of these into the first original equation, we have to square it. We’ll

get the same thing in both cases, so we might as well substitute directly for x^2 into the first

original equation to obtain

y=a 1 (b 2 −b 1 ) / (a 1 −a 2 )+b 1

Multiplying the right-hand side out, we get

y= (a 1 b 2 −a 1 b 1 ) / (a 1 −a 2 )+b 1

Expressing b 1 as the ratio b 1 /1, applying the general rule for adding ratios, and then canceling

out identical terms that subtract from each other, we obtain

y= (a 1 b 2 −a 2 b 1 ) / a 1 a 2

Writing these solutions as ordered pairs is tricky and messy. We might as well state the x and y

values separately and take advantage of the plus-or-minus sign. Then we can write

x=±[(b 2 −b 1 ) / (a 1 −a 2 )]1/2 and y= (a 1 b 2 −a 2 b 1 ) / a 1 a 2

Question 27-5

What happens when we plug the solution for x^2 into the second original equation in the above

system (instead of the first one, which we already did) and solve for y?

Answer 27-5

When we substitute directly for x^2 into the second original equation, we get

y=a 2 (b 2 −b 1 ) / (a 1 −a 2 )+b 2

Multiplying the right-hand side out, we get

y= (a 2 b 2 −a 2 b 1 ) / (a 1 −a 2 )+b 2

Expressing b 2 as the ratio b 2 /1, applying the general rule for adding ratios, and then canceling

out identical terms that subtract from each other, we obtain the same result as before:

y= (a 1 b 2 −a 2 b 1 ) / a 1 a 2

Question 27-6

Consider the following pair of quadratic equations as a two-by-two system:

y= (x+ 1)^2

and

y= (x− 1)^2

How can we find the real solutions of this system?