Multiplying and Dividing Rational Expressions

[malelia123]

Hey Guys - I've got a question about the process of multiplying rational expressions
I'm just not sure what steps to take when solving for the following (of Course the textbook shows nothing on this)

3x(2x-1) / 7x^2 (x-6)

multiplied by

14(6-x) / 8x^3

The parenthesis in the equations are what is screwing me up.

I tired to solve the items in parenthesis first but my answer came out wrong.

If you can help show me the process/steps for simplifying this complicated rational expressions it would be greatly appreciated!

Thank you

 

[JeffM]

Hey Guys - I've got a question about the process of multiplying rational expressions
I'm just not sure what steps to take when solving for the following (of Course the textbook shows nothing on this)

3x(2x-1) / 7x^2 (x-6)

multiplied by

14(6-x) / 8x^3

The parenthesis in the equations are what is screwing me up.

I tired to solve the items in parenthesis first but my answer came out wrong.

If you can help show me the process/steps for simplifying this complicated rational expressions it would be greatly appreciated!

Thank you

First of all I am fairly confident that you miswrote the first expression

What you wrote is \(\displaystyle 3x * \dfrac{2x - 1}{7} * x^2 * (x - 6).\)

What I suspect you meant was \(\displaystyle \dfrac{3x(2x - 1)}{7x^2(x - 6)}.\) Remember PEMDAS when you are writing out your problems.

The basic idea is to gather like terms first and then to cancel as many common factors as possible before doing any complicated multiplication.

\(\displaystyle \dfrac{3x(2x - 1)}{7x^2(x - 6)} * \dfrac{14(6 - x)}{8x^3} = \dfrac{3 * 14 * x(2x - 1)(6 - x)}{7 * 8 * x^5(x - 6)}.\) Gather like terms.

\(\displaystyle \dfrac{3 * 14 * x(2x - 1)(6 - x)}{7 * 8 * x^5 *(x - 6)} = \dfrac{3 * 2 * (2x - 1) * (- 1)(x - 6)}{8x^4(x - 6)} = \dfrac{- 3 * 2 *(2x - 1)}{2 * 4x^4} = - \dfrac{3(2x - 1)}{4x^4}.\) Cancel where you can

Now check If x = 2

\(\displaystyle \dfrac{3 * 2(2 * 2 - 1)}{7* 2^2(2 - 6)} * \dfrac{14(6 - 2)}{8 * 2^3} = \dfrac{2 * (3 * 3) * 2 * 7 (4)}{7 * 4 * (- 4) * 8 * 8} = - \dfrac{7 * 16 * 9}{7 * 16 * 64} = - \dfrac{9}{64} = \dfrac{3(3)}{4 * 16} = \dfrac{3(2 * 2 - 1)}{4 * 2^4}.\) It checks

 

[malelia123]

First of all I am fairly confident that you miswrote the first expression

What you wrote is \(\displaystyle 3x * \dfrac{2x - 1}{7} * x^2 * (x - 6).\)

What I suspect you meant was \(\displaystyle \dfrac{3x(2x - 1)}{7x^2(x - 6)}.\) Remember PEMDAS when you are writing out your problems.

The basic idea is to gather like terms first and then to cancel as many common factors as possible before doing any complicated multiplication.

\(\displaystyle \dfrac{3x(2x - 1)}{7x^2(x - 6)} * \dfrac{14(6 - x)}{8x^3} = \dfrac{3 * 14 * x(2x - 1)(6 - x)}{7 * 8 * x^5(x - 6)}.\) Gather like terms.

\(\displaystyle \dfrac{3 * 14 * x(2x - 1)(6 - x)}{7 * 8 * x^5 *(x - 6)} = \dfrac{3 * 2 * (2x - 1) * (- 1)(x - 6)}{8x^4(x - 6)} = \dfrac{- 3 * 2 *(2x - 1)}{2 * 4x^4} = - \dfrac{3(2x - 1)}{4x^4}.\) Cancel where you can

Now check If x = 2

\(\displaystyle \dfrac{3 * 2(2 * 2 - 1)}{7* 2^2(2 - 6)} * \dfrac{14(6 - 2)}{8 * 2^3} = \dfrac{2 * (3 * 3) * 2 * 7 (4)}{7 * 4 * (- 4) * 8 * 8} = - \dfrac{7 * 16 * 9}{7 * 16 * 64} = - \dfrac{9}{64} = \dfrac{3(3)}{4 * 16} = \dfrac{3(2 * 2 - 1)}{4 * 2^4}.\) It checks

Thanks Guys! That was exactly the answer I was looking for, It's been so many years since I took Grade 11 Math I'm having a hard time remember all the little steps.

I really appreciate it!