please help with solving (2ax/4b) + (3c/2d) - x = 1 for x

[atucker9]

someone please help me... im helping my brother with his online homework and i just graduated from college and you would think i would know how to do this but im just stuck... it says solve for x and the problem is:

(2ax/4b) + (3c/2d) - x = 1

if you dont want to show me all the work and the final answer can you at least lead me in the right direction....

 

[Loren]

Re: please help with solving for x

That means to get one x all by itself on one side of the equation with everything else with no x's on the other side.

(2ax/4b) + (3c/2d) - x = 1

\(\displaystyle \frac{2ax}{4b}+\frac{3c}{2d}-x=1\)

It's usually a good idea to eliminate fractions by multiplying both sides by the least common denominator.
\(\displaystyle 4bd(\frac{2ax}{4b}+\frac{3c}{2d})-4bdx=4bd\)

\(\displaystyle 2adx + 6bc - 4bdx = 4bd\)
\(\displaystyle 2adx -4bdx = 4bd -6bc\)
\(\displaystyle x(2ad-4bd)=4bd-6bc\)
\(\displaystyle x=\frac{4bd-6bc}{2ad-4bd}=\frac{2b(2d-3c)}{2d(a-2b)}=\frac{b(2d-3c)}{d(a-2b)}\)

Hope I didn't goof.

 

[soroban]

Re: please help with solving for x

Hello, atucker9!

Loren is absolutely correct . . . (I would reduce first.)

\(\displaystyle \text{Solve for }x\!:\;\;\frac{2ax}{4b} + \frac{3c}{2d} - x \;=\;1\)

\(\displaystyle \text{Reduce first: }\;\frac{ax}{2b} + \frac{3c}{2d} - x \;=\;1\)

\(\displaystyle \text{Multiply by the LCD, }2bd\!:\quad 2bd\left(\frac{ax}{2b} + \frac{3c}{2d} - x \;=\;1\right)\)

. . \(\displaystyle \text{and we have: }\;adx + 3bc - 2bdx \;=\;2bd \quad\Rightarrow\quad adx - 2bdx \;=\;2bd-3bc\)


\(\displaystyle \text{Factor: }\;d(a-2b)x \;=\;b(2d-3c)\)


\(\displaystyle \text{Therefore: }\;x \;=\;\frac{b(2d-3c)}{d(a-2b)}\)

 

[Loren]

Of course. It is usually a good idea to reduce a fraction first.