Help on my IB Higher Level Math Summer Assignment?

Question

I need to simplify:

[ ( a / b ) - a ] / [ a + ( a / b ) ^2 ]

so that it equals:

[ b ( 1 - b ) ] / [ b^2 + a ]

Please help, and show working with explanation if necessary.

Answer 1

Are you sure this is IB HL Math? Seems rather simplistic...

Well no matter,

Let's take the problem half at a time.

First take a look at the numerator, you want that expression to have a common denominator. This will give you:

(a-ab)/b

Then do the same with the denominator, evaulating the exponent first:

(ab^2+a^2)/b^2

Now remember that when you divide by a fraction it is the same as multiplying as it's reciprocal?

so the whole nasty division problem becomes a easier multiplication one:

(a-ab)/(b) * (b^2)/(ab^2+a^2)

Cancel a b and you get:

(ab-ab^2)/(ab^2+a^2)

Now you can factor out an a from both the top and bottom, canceling it:

(b-b^2)/(b^2+a)

Hey! We're almost there! The bottom matches! Now simply factor out a b and voila!

[ b ( 1 - b ) ] / [ b^2 + a ]

Answer 2

((a-ba)/b) / (b^2 a+a^2)/b^2)= make b a common denominator on the left. On the right expand the fraction part to a^2/ b^2 then combine the fraction with a using b^2 as a common denominator.

next divide the fractions by multiplying the reciprocal of the second fraction. then reduce terms a & b common to both to get:

[a(1-b)]/b * [b^2 / [a(b^2+a)]
(1-b)(b)/(b^2+a)

Answer 3

[(a/b) - a] / [ a + (a/b)²]
= [(a/b) - a] / [ a + (a²/b²)]

You must first take care of the fractions above and below the main fraction - Get it so they have same denominators.
Then you can add them together still keeping the main fraction present though.

= [[(a/b) - a] * (b/b)] / [[ a + (a²/b²)] * (b²/b²)]
= [(a/b) - (ab/b)] / [(ab²/b²) + (a²/b²)] --> Add together.
= [(a - ab)/b] / [(ab² + a²)/b²]

Now, get it so the whole fraction has the same denominator.
[(a - ab)/b] / [(ab² + a²)/b²] * (b²/b²)

= [(a - ab)/b]*(b²) / [(ab² + a²)/b²]*(b²)
= [(ab² - ab³)/b] / [(ab^4 + a²b²)/b²) -->Reduce b's.
= (ab - ab²) / (ab² + a²) -->Reduce by common factors.

= [ab(1-b)] / [a(b²+a)] -->Divide what's on both top & bottom. Which is "a", not in the brackets.

= b(1-b) / (b²+a)

Answer 4

[ ( a / b ) - a ] / [ a + ( a / b ) ^2 ]
Putting [ ( a / b ) - a ] over common denominator b gives:
(a - ab) / b .........(1)
Putting [ a + ( a / b ) ^2 ] over common denominator b^2 gives:
(ab^2 + a^2) / b^2 ........(2)
Turn (2) upside down, and multiply by (1):
(a - ab) b^2 / b(ab^2 + a^2)
Factor a out of each bracket:
a(1 - b)b^2 / ab(b^2 + a)
Cancel ab from numerator and denominator:
b(1 - b) / (b^2 + a).