How do you solve #\frac { 3} { x } + \frac { x - 1} { 2x } = 1#?

3 Answers
Dec 9, 2016

#x = 5#

Explanation:

First, we must get each fraction over a common denominator so we can add them by multiplying by the necessary form of #1#:

#(2)/(2)(3/x) + (x - 1)/(2x) = 1#

#6/(2x) + (x - 1)/(2x) = 1#

#(6 + x - 1)/(2x) = 1#

#(5 + x)/(2x) = 1#

Now, we can multiply each side of the equation by #2x# to eliminate the fraction while keeping the equation balanced:

#(2x)(5 + x)/(2x) =1*2x#

#cancel((2x))(5 + x)/cancel(2x) =2x#

#5 + x = 2x#

We can now solve for #x# while keeping the equation balanced:

#5 + x - x = 2x - x#

#5 + 0 = (2 - 1)x#

#5 = 1x#

#5 = x#

#x = 5#

Dec 9, 2016

# x= 5#

Explanation:

Using short cuts

Write as #(x-1)/(2x)=1-3/x#

#(x-1)/(2x)=(x-3)/x#

Cross multiply

#x(x-1)=2x( x-3) #

#x^2-x=2x^2-6x#

#x^2-5x=0#

#x(x-5)=0#

#x=0 and x= 5#

But for #x=0# we have:

#3/0+...#

There is no need to consider the rest as divide by 0 is 'not permitted'. It is an excluded value.

So the answer is just #x=5#

Mar 28, 2017

#5 =x#

Explanation:

When you have an equation that has fractions, you can get rid of the denominators immediately.

Multiply each term by the LCM of the denominators, so you can cancel them.

In this case the LCM is #color(red)(2x)#

#(color(red)(2x)xx3)/x +(color(red)(2x)xx(x-1))/(2x) = 1xxcolor(red)(2x)#

#((2cancelx)xx3)/cancelx +(cancel(2x)xx(x-1))/cancel(2x) = 1xx(2x)#

#6+x-1 =2x#

#5 = 2x-x#

#5 =x#