SOLUTION: I do not get how to do this problem when it comes to fractions. X = 2X/5 + 1 I know you have to combine like terms so you somehow move 2x/5 to the other side of the equal sign, c

Algebra ->  Linear-equations -> SOLUTION: I do not get how to do this problem when it comes to fractions. X = 2X/5 + 1 I know you have to combine like terms so you somehow move 2x/5 to the other side of the equal sign, c      Log On


   



Question 4309: I do not get how to do this problem when it comes to fractions.
X = 2X/5 + 1
I know you have to combine like terms so you somehow move 2x/5 to the other side of the equal sign, correct?

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Here are two methods to solve this problem. You decide which one is easier for you. The first method is for everyone who hates fractions, and wants to get rid of them in the first step. Begin by multiplying both sides by the common denominator to eliminate the fractions.
x+=+2x%2F5+%2B+1

Multiply each side by 5:
5%2Ax+=+5%2A+%282x%29%2F5+%2B+5+%2A1

(This next step is VERY important!)
Divide out the 5. Do NOT multiply 5 times 2x!!
5x+=+2x+%2B+5

Next, subtract 2x from each side:
5x+-+2x+=+2x+-+2x+%2B+5
3x+=+5

Divide each side by 3, and you get
+3x%2F3+=+5%2F3
x+=+5%2F3

It doesn't come out even, but then life usually doesn't!!


Another way to solve this x+=+2x%2F5+%2B+1 is to not even worry about the denominators, and just subtract 2x%2F5 from each side of the equation. This gives you:
x+-+2x%2F5+=+2x%2F5+-+2x%2F5+%2B+1
3x%2F5+=+1

Now multiply both sides by the reciprocal of 3%2F5 which is 5%2F3, and you get:
5%2F3+%2A+3x%2F5+=+5%2F3+%2A+1
x+=+5%2F3

R^2 at SCC