# Thread: Equation with decimal problem

1. ## Equation with decimal problem

0.05x-2 = 0.3(x-5)

5x-2=3x-15

5x=3x-13

2x/2=-13/2

x=-13/2

I keep getting this answer, but in the book states its x=-2 which is correct but I can't figure how?

Thank you

2. Originally Posted by Aristoxenus
0.05x-2 = 0.3(x-5)

5x-2=3x-15

5x=3x-13

2x/2=-13/2

x=-13/2

I keep getting this answer, but in the book states its x=-2 which is correct but I can't figure how?

Thank you
I'm not sure what you wanted to get by your 2nd step (?). I assume you want to get rid of the decimals (?). If so:

1. Multiply through by 100:

$\displaystyle 0.05x-2 = 0.3(x-5)~\implies~5x-200=30(x-5)$

2. Expand the bracket and collect like terms:

$\displaystyle 5x-200=30x-150~\implies~-25x = 50$

Solve for x.

3. O, I see you multiply both sides by 100. I was multiplying 0.05 by 100 and the right side of the equation 0.3 by 10 So im assuming you must keep the continuity by multplying both sides by 100.

Thank you

4. Originally Posted by Aristoxenus
O, I see you multiply both sides by 100. I was multiplying 0.05 by 100 and the right side of the equation 0.3 by 10 So im assuming you must keep the continuity by multplying both sides by 100.

Thank you
Yes, whenever you do something to the left side of an equation, you must do the same thing to the other side.

But be careful, that wasn't your only problem- when you multiply the left side by 100, you multiply the entire left side- and you didn't do that before. 100(.05x- 2)= 100(.05x)- 100(2)= 5x- 200. That is why earboth has "200" where you had only "2".