Help with steps solving for x

[christobel]

Hi there,
Although I have a basic understanding of algebra I am having trouble solving the following problem.

(0.2 - x)/1.2 = (0.16 - x)/0.9

Could someone please explain the steps and order you would take to solve for x

NOTE & WORKINGS: This is out of a text book Fundamentals of corporate accounting by Ross et al pg 352 question 11.2. It actually gives you the answer but I don't understand the order of workings. They have written the following:
(0.2 - x) = (0.16 - x)(1.2/0.9)
0.2 - 0.16 X (4.3) = x - x X (4/3)
x = 0.04

Where I am stuck is when you times both sides by 1.2, why is the 0.9 not also times by 1.2

Thanks in advance

 

[pka]

Hi there,
Although I have a basic understanding of algebra I am having trouble solving the following problem.
(0.2 - x)/1.2 = (0.16 - x)/0.09

\(\displaystyle (0.09(0.2-x)=(1.2)(0.16-x)\)

 

[Quaid]

(0.2 - x)/1.2 = (0.16 - x)/0.09

Did you intend to type 0.09 above?

This is out of a text book

They have written the following:

x = 4

I don't understand their work, and their solution is not correct.

Where I am stuck is when you times both sides by 1.2, why is the 0.9 not also times by 1.2

The numerator on each side gets multiplied; 0.9 is in the denominator.

Please double-check the typing in your post.

 

[christobel]

\(\displaystyle (0.09(0.2-x)=(1.2)(0.16-x)\)

Thank you, I did try this way but again gone stuck..... could you please let me know the next step you took

 

[mmm4444bot]

Please show your work.

 

[christobel]

(0.2 - x)/1.2 = (0.16 - x)/0.9
= (0.9)(0.2 - x)= (1.2)(0.16 - x)
= 0.18 - 0.9x = 0.192 - 1.2x
= 0.18 - 0.9x + 1.2x = 0.192
= 0.3x = 0.012
= x = 0.04

Thank you!!! I had put an extra 0 in before the .9

Sometimes when you work on a problem for so long you are blinded by the entries