Question

What is `1/(v-1) -: (9v^2 - 63v)/(v^2-7v+6)`?

Answer 1

You must first flip the second fraction, to transform the expression into a multiplication.

Explanation:

`1/(v - 1) xx (v^2 - 7v + 6)/(9v^2 - 63v)`

We now must factor everything completely to see what we can eliminate before multiplying.

`1/(v - 1) xx ((v - 6)(v - 1))/(9v(v - 7)`

The (v - 1)'s cancel themselves out. We are left with: `(v - 6)/(9v(v - 7))`

That is quite simple to do. All you need is to master all your factoring techniques. However, now we must identify non-permissible values for x. This becomes slightly tricky with divisions. Inspect the following rational expression.

`(2x)/(x^2 + 6x + 5)`

What values are non-permissible for x?

For this, you must set the denominator to 0 and solve for x.

`x^2 + 6x + 5 = 0`

`(x + 5)(x + 1) = 0`

`x = -5 and -1`

So, x cannot be -5 or -1. The reason for this is that it makes the denominator 0, and division by 0 is non defined in mathematics.

Back to your problem. In a division, its more complicated. You must account for all possible denominators.

Scenario 1:

`v - 1 = 0`

`v = 1`

So, we already know v cannot be equal to 1.

Scenario 2:

`v^2 - 7x + 6 = 0`

`(v - 6)(v - 1) = 0`

`v = 6 and v = 1`

So, we now know v cannot be 6 or 1.

Scenario 3 (since the numerator of the second expression becomes the denominator when you transform the operation into a multiplication, you have to find any NPV's here as well):

`9v^2 - 63v = 0`

`9v(v - 7) = 0`

`v = 0 et 7`

In summary, our non permissible values are x = 0, 1, 6, and 7.

Practice exercises:

Divide and simplify completely. State all non permissible values.

`(10x^2 + 42x + 36)/(6x^2 - 2x - 60) -: (40x + 48)/(3x^2 - 13x + 10)`