Need help

lev888

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Let's take #4: 7=4/n
You multiplied both sides by 4 and got 28 = n.
28 is correct but n is not. Let's say n is 1. (4/1)*4 = 4*4 = 16. Definitely not 1.
Could you look up how to multiply a fraction by a number?
 

Dr.Peterson

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Both of the problems you ask about have a variable in the denominator. A key step will be to multiply both sides by the denominator, in part to get the variable out of there. There's more to do here than just eliminate a coefficient.

But you are also doing things that don't accomplish what you claim. Does \(\displaystyle \frac{-4}{3k}\cdot-\frac{4}{3} = k\) as you claim? Write it out this way, rather than in a column, and think carefully about it.
 

Narwalalpaca12$

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Both of the problems you ask about have a variable in the denominator. A key step will be to multiply both sides by the denominator, in part to get the variable out of there. There's more to do here than just eliminate a coefficient.

But you are also doing things that don't accomplish what you claim. Does \(\displaystyle \frac{-4}{3k}\cdot-\frac{4}{3} = k\) as you claim? Write it out this way, rather than in a column, and think carefully about it.
Could you give me an example please?
 

lev888

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lev888

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Ok.
Let's try to solve for x formally. This means starting with 4=8/x get to x=something.
Step 1. How do we get x from the world of denominators to the numerators? We multiply both sides by x: 4*x = (8/x)*x => 4x = 8*(x/x) => 4x = 8
Step 2. How do we get x out of 4x? Divide both side by 4: (4x)/4 = 8/4 => (4/4)*x = 2 => x = 2
Makes sense?
 

Narwalalpaca12$

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So would I do this...
1. 4*k= (3/k)*k = 4K/3k=8
2.4k/3k divided by 4/3 is k/k and 8 divided by 4/3 = 6
 

lev888

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So would I do this...
1. 4*k= (3/k)*k = 4K/3k=8
2.4k/3k divided by 4/3 is k/k and 8 divided by 4/3 = 6
Rule number 1: never forget about signs! A huge number of errors in algebra are caused by incorrect handling of signs.
The equation is: -4/(3k) = 8
We are multiplying both sides by k. Left hand side is -4/(3k). Right hand side is 8. Can you show the result? What you wrote above in 1. is not correct.
 

Narwalalpaca12$

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I am very confused
 

Narwalalpaca12$

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I didn’t understand what you did with the first step
 

Dr.Peterson

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Let's look at \(\displaystyle 4 = \frac{8}{x}\).

We want to solve for x; but it is in the bottom of a fraction, which is not good. In order to undo that division by x, we multiply both sides by x. Since the right side is a fraction, it will be helpful to write x as the fraction x/1:

\(\displaystyle 4\cdot x = \frac{8}{x}\cdot\frac{x}{1}\)​

When we do that multiplication, the x cancels out:

\(\displaystyle 4x = \frac{8}{1}\)​

So we have \(\displaystyle 4x = 8\).

Now, we want to get x by itself, so we have to undo the multiplication by 4; we do that by dividing by 4:

\(\displaystyle \frac{4x}{4} = \frac{8}{4}\)​

\(\displaystyle x = 2\)​

If you don't follow that, please tell us, again, where you are having trouble, and why.
 
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