Algebra Know-It-ALL

The ratio of the nonnegative square root of 50 to the nonnegative square root of 2 is

exactly equal to 5, even though the dividend and the divisor are both irrational.

8. Here’s the sum-of-quotients rule again:

w/x+y/z= (wz+xy)/(xz)

In this example, we can let w= 7, x= 11, y=−5, and z= 17. None of these are equal to 0,

so we can be sure the rule will work properly. Now it’s simply a matter of doing the

arithmetic:

7/11+ (−5/17)= {(7 × 17) + [11 × (−5)]} / (11 × 17)

= [119 + (−55)] / 187

= 64 / 187

This can’t be reduced to lower terms because the denominator, 187, is the product of

two primes, 11 and 17. Neither of these factors “goes into” the numerator, 64, without

leaving a remainder.

9. To solve this, we can rewrite (x−y) as [x+ (−y)]. Then, using the product of sums rule,
   we have

(x+y)[x+ (−y)]=xx+x(−y)+yx+y(−y)

We can use familiar arithmetic rules to write this as

=x^2 −xy+xy−y^2

The addends −xy and xy cancel out here, so we get the final result

=x^2 −y^2

10. Letu,v,w,x,y, and z be real numbers. We’re given a product of sums of three variables,
    and we’re told to multiply it out using the product of sums rule. That rule, as stated in
    the chapter text, only allows us to use two variables in each addend. But we can “cheat”
    by renaming certain sums! Here is the original expression:

(u+v+w)(x+y+z)

Let’s rename (v+w) as r, and (y+z) as s.Then we have

(u+r)(x+s)

Using the product of sums rule, we can multiply this out to

ux+us+rx+rs

Now let’s substitute the original values for r and s back into the expression. This gives us

ux+u(y+z)+ (v+w)x+ (v+w)(y+z)

Chapter 9 617