LCD? Equations and inequalities

[Blackwinter]

Homework Statement

1/z - 1/2z - 1/5z = 10/(z+1)

This is the equation presented right after in the textbook where the author explains how to use the LCD to convert this equation into a more understood linear equation.

Homework Equations

Here is the example before this equation:
1/(1+x) + 1/(x-2) = x+3/(x^2-x-2)

The LCD is (x+1)(x-2)

This is multiplied in and then we have: (x-2)+(x+1)=x+3
then we simplify it to x=4

The Attempt at a Solution

I am having troubles, I am not sure what the LCD would be, my current thoughts are:
10z(z+1)

Please ask, if I left anything out.

 

[scurty]

Homework Statement

1/z - 1/2z - 1/5z = 10/(z+1)

This is the equation presented right after in the textbook where the author explains how to use the LCD to convert this equation into a more understood linear equation.

Homework Equations

Here is the example before this equation:
1/(1+x) + 1/(x-2) = x+3/(x^2-x-2)

The LCD is (x+1)(x-2)

This is multiplied in and then we have: (x-2)+(x+1)=x+3
then we simplify it to x=4

The Attempt at a Solution

I am having troubles, I am not sure what the LCD would be, my current thoughts are:
10z(z+1)

Please ask, if I left anything out.

That would be correct. The process I use is to split every term in the denominators into prime factors. For polynomials this would mean factoring them. The LCM would be the product of the greatest occurence of each prime factor in each fraction's denominators. Say you had 24 and 18. This would become $2 \cdot 2 \cdot 2 \cdot 3$ and $2 \cdot 3 \cdot 3$. The greatest amount of 2s would be three in the first umber and the greatest amount of 3s would be two in the second number. Therefore the LCM is $2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 = 72$.

I don't know exactly what you are having trouble with, so I hope that cleared up any issues.

 

[Blackwinter]

So, I have tried this question several times and have yet to succeed in doing it. I just wanted to make sure my conclusions about the LCD were correct, than I would proceed from there and see how I'd do.

I am currently multiplying 10z(z+1) into the original equation. I am unsure what 10z(z+1) multiplied into 10 is. Is it: 100z(z+1)?

Here's my full attempt at the problem:I multiplied 10z(z+1) into 1/z - 1/2z - 1/5z = 10/(z+1) getting:

10z(z+1)/z - 10z(z+1)/2z - 10z(z+1)/5z = 100z(z+1)/(z+1)

This in my mind simplifies to:

9z(z+1) - 8z(z+1) - 5z(z+1) = 100z

If that is right, I continued to then simplify:
-4z(z+1) = 100z

This is where my low quality brain is most confused:

My attempts don't seem to prevail here and I suspect I erred earlier. I have thoughts of multiplying the -4z into (z+1) and then factoring that, but I'm not sure.

Hopefully I communicated everything well.Thank you so much!

 

[Blackwinter]

Linear equations.

Homework Statement

1/z - 1/2z - 1/5z = 10/(z+1)

This is the equation presented in the textbook right after where the author explains how to use the LCD to convert this equation into a more understood linear equation.

Homework Equations

Here is the example before this equation:
1/(1+x) + 1/(x-2) = x+3/(x^2-x-2)

The LCD is (x+1)(x-2)

This is multiplied in and then we have: (x-2)+(x+1)=x+3
then we simplify it to x=4

The Attempt at a Solution

So, I have tried this question several times and have yet to succeed in doing it. I just wanted to make sure my conclusions about the LCD were correct, than I would proceed from there and see how I'd do.

I am currently multiplying 10z(z+1) which is the LCD into the original equation. I am unsure what 10z(z+1) multiplied into 10 is. Is it: 100z(z+1)?

Here's my full attempt at the problem:


I multiplied 10z(z+1) into 1/z - 1/2z - 1/5z = 10/(z+1) getting:

10z(z+1)/z - 10z(z+1)/2z - 10z(z+1)/5z = 100z(z+1)/(z+1)

This in my mind simplifies to:

9z(z+1) - 8z(z+1) - 5z(z+1) = 100z

If that is right, I continued to then simplify:
-4z(z+1) = 100z

This is where my low quality brain is most confused:

My attempts don't seem to prevail here and I suspect I erred earlier. I have thoughts of multiplying the -4z into (z+1) and then factoring that, but I'm not sure.

Hopefully I communicated everything well.


Thank you so much!

 

[lurflurf]

all is good up to here
10z(z+1)/z - 10z(z+1)/2z - 10z(z+1)/5z = 100z(z+1)/(z+1)
you have subtracted where you should have divided
10/1=10 not 9
10/2=5 not 8
10/5=2 not 5
also you have not canceled correctly the variables
z(z+1)/z=(z+1) not z(z+1)
another approach would be to add the like terms
$$\frac{1}{z}-\frac{1}{2z}-\frac{1}{5z}=\frac{10}{z+1}\\
\left( 1-\frac{1}{2}-\frac{1}{5} \right)\frac{1}{z}=\frac{10}{z+1}$$

 

[berkeman]

Two threads merged...

 

[Blackwinter]

What do you mean by: "also you have not canceled correctly the variables
z(z+1)/z=(z+1) not z(z+1)"

Sorry, I have many holes in my math (Extended period away from it.)

1- 1/2 - 1/5 = 7/10?

 

[Mark44]

What do you mean by: "also you have not canceled correctly the variables
z(z+1)/z=(z+1) not z(z+1)"

$$\frac{z(z + 1)}{z} = \frac{z}{z}\cdot\frac{z + 1}{1}= z + 1$$
z/z = 1, so can be removed.
Cancelling means finding factors that are the same in both numerator and denominator.
Also, "factors" implies that both numerator and denominator appear as products.

Sorry, I have many holes in my math (Extended period away from it.)

1- 1/2 - 1/5 = 7/10?

No, pretty far off. Here the LCD is 10, so multiply each term by 1 in some form so as to get a denominator of 10.

$$1\frac{10}{10} - \frac{1}{2}\frac{5}{5} - \frac{1}{5}\frac{2}{2}$$
$$= \frac{10}{10} - \frac{5}{10} - \frac{2}{10}$$
Now that all the denominators are the same, the fractions can be added to get a single fraction with 10 in the denominator.