Can someone help me with 2/3X + 8 = 1/4X ?
- Thread starter jmullett
- Start date
D
Deleted member 4993
Guest
Re: Can someone help me with this problem?
What do you need to with the given expression?
If your task is to solve for 'X' (then you should have said so) -
can you solve for "X" - given
2X +8 = 3X..........................?
If you don't know - how to do above then go to:
http://www.purplemath.com/modules/solvelin.htm
for worked out examples. Review those then
please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
jmullett said:
What do you need to with the given expression?
If your task is to solve for 'X' (then you should have said so) -
can you solve for "X" - given
2X +8 = 3X..........................?
If you don't know - how to do above then go to:
http://www.purplemath.com/modules/solvelin.htm
for worked out examples. Review those then
please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
I disagree with staple.
Following the order of operations 2/3X + 8 = 1/4X means \(\displaystyle \frac{2}{3}x + 8 = \frac{1}{4}x\) or \(\displaystyle \frac{2x}{3} + 8 = \frac{x}{4}\).
2/(3X) + 8 = 1/(4X) means \(\displaystyle \frac{2}{3x} + 8 = \frac{1}{4x}\).
To solve for x start by multiplying both sides of the equation by the least common denominator of the fractions.
Following the order of operations 2/3X + 8 = 1/4X means \(\displaystyle \frac{2}{3}x + 8 = \frac{1}{4}x\) or \(\displaystyle \frac{2x}{3} + 8 = \frac{x}{4}\).
2/(3X) + 8 = 1/(4X) means \(\displaystyle \frac{2}{3x} + 8 = \frac{1}{4x}\).
To solve for x start by multiplying both sides of the equation by the least common denominator of the fractions.