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Simple rearranging

So an idiot forgot how to rearrange for x only, any help would be appreciated.

6.31\mathrm x 10^{-6}=\dfrac{1.78\mathrm x 10^{-4} \times \frac{1}{10} -4x}{4x}

which is the original thing i had

So far i have gotten to

\dfrac {4x}{\frac{1}{10} -4x} = \dfrac{1.78\mathrm x 10^{-4}}{6.31\mathrm x 10^{-6}}

Help pls.

Sorry, i think i have it.... sorry for wasting my time and anyone else's who looks at this :/

(edited 7 years ago)

Reply 1

nerak99

OK, you have a fraction equals a fraction so you can cross multiply.

a/b=c/d >>>>>ad=bc

Which gets rid of the fractions and should enable you to progress.

This assumes that there are really brackets around the 1/10 -4x

Reply 2

will'o'wisp

Original post by nerak99

There are brackets around the 1/10 term i don't get what u mean but i've done it.

Reply 3

CraigFowler

Did your answer look like this?

Photo.jpg175.2KB

Reply 4

nerak99

"i don't get what u mean but i've done it"

Cross multiplying is like this. You have a situation where fractions are making life look difficult. In this case, there is a fraction on the left and one on the right. A common scenario like yours.

\frac{\text{bit of algebra a}}{\text{bit of algebra b }}=\frac{\text{bit of algebra c}}{\text{bit of algebra d }}

You can multiply through by algebra d and also algebra b which brings algebra b to the RHS and algebra d to the LHS

{\text{bit of algebra a}} \times {\text{bit of algebra d }}={\text{bit of algebra c}} \times {\text{bit of algebra b }}

Hurray, all the fraction stuff is gone. I have always known this as cross multiplying.

Only works if you have a single fraction on the left = a single fraction on the right. This will not work if you had (say)

\frac{\text{bit of algebra a}}{\text{bit of algebra b }}+ y=\frac{\text{bit of algebra c}}{\text{bit of algebra d }}

Multiply through by (algebra b) X (algebra d)

or

\frac{\text{bit of algebra a}}{\text{bit of algebra b }} + \frac{\text{bit of algebra g}}{\text{bit of algebra h }}=\frac{\text{bit of algebra c}}{\text{bit of algebra d }}

Multiply through by (algebra b) X (algebra d) X (algebra h)

(edited 7 years ago)

Reply 5

CraigFowler

Original post by nerak99

\frac{\text{bit of algebra a}}{\text{bit of algebra b }}=\frac{\text{bit of algebra c}}{\text{bit of algebra d }}

You can multiply through by algebra d and also algebra b which brings algebra b to the RHS and algebra d to the LHS

{\text{bit of algebra a}} \times {\text{bit of algebra d }}={\text{bit of algebra c}} \times {\text{bit of algebra b }}

\frac{\text{bit of algebra a}}{\text{bit of algebra b }}+ y=\frac{\text{bit of algebra c}}{\text{bit of algebra d }}

Multiply through by (algebra b) X (algebra d)

or

\frac{\text{bit of algebra a}}{\text{bit of algebra b }} + \frac{\text{bit of algebra g}}{\text{bit of algebra h }}=\frac{\text{bit of algebra c}}{\text{bit of algebra d }}

Multiply through by (algebra b) X (algebra d) X (algebra h)

I have checked the answer 0.02415 and it is correct.

One can cross multiply with fraction and whole number when one uses the fact that the whole number can be expressed as a fraction, e.g. 4 becomes 4/1 and 5 becomes 5/1 etc.

In the examples 1/10 - 4x becomes 1-40x / 10

Reply 6

nerak99

Original post by CraigFowler

You are correct that the denominators like "bit of algebra b" could be as simple as just 1.

(edited 7 years ago)

Reply 7

will'o'wisp

Original post by nerak99

\frac{\text{bit of algebra a}}{\text{bit of algebra b }}=\frac{\text{bit of algebra c}}{\text{bit of algebra d }}

You can multiply through by algebra d and also algebra b which brings algebra b to the RHS and algebra d to the LHS

{\text{bit of algebra a}} \times {\text{bit of algebra d }}={\text{bit of algebra c}} \times {\text{bit of algebra b }}

\frac{\text{bit of algebra a}}{\text{bit of algebra b }}+ y=\frac{\text{bit of algebra c}}{\text{bit of algebra d }}

Multiply through by (algebra b) X (algebra d)

or

\frac{\text{bit of algebra a}}{\text{bit of algebra b }} + \frac{\text{bit of algebra g}}{\text{bit of algebra h }}=\frac{\text{bit of algebra c}}{\text{bit of algebra d }}

Multiply through by (algebra b) X (algebra d) X (algebra h)

whoops derp i know what cross multiplying sorry for making u type all that out :/

Thanks anyway!

Reply 8

nerak99

What does "Whoops Derp" mean?

Reply 9

College rocks

Can someone please help to make x the subject:

(244.2 - X)/X = 0.418

What would be the value of X?

Reply 10

mathcoachni

Original post by College rocks

Can someone please help to make x the subject:

(244.2 - X)/X = 0.418

What would be the value of X?

I haven't got a calculator handy...so I'll leave the final solution to you.
Step 1: multiply both sides by x. Gives 244.2-x = 0.418x

Step 2 add x to both sides. Gives 244.2 =1.418x
Step 3 find x by dividing both sides by 1.418 👍