help with this please

[Tresa332006]

Can any one show me another way to work this problem:
(x - 4)/2 - 1/5 = (7x + 1)/20

I tried moving (7x + 1)/20 to the left of the equal sign and I got lost. Was I right or am I doing the problem all wrong.

 

[Mrspi]

Can any one show me another way to work this problem:
(x - 4)/2 - 1/5 = (7x + 1)/20

I tried moving (7x + 1)/20 to the left of the equal sign and I got lost. Was I right or am I doing the problem all wrong.

Since this is an equation, you can multiply both sides by the same number. If you choose a multiplier that all of the denominators will divide into, you'll end up with an equation with NO fractions (how neat is that?!).

Since 2, 5, and 20 will all divide evenly into 20, let's multiply both sides of the equation by 20:

20* (x - 4)          20*( 1 )        20 * (7x + 1)
    ------      -       -----     =       ---------
       2                  5                  20

Divide each denominator into the multiplier, and this is what you should have left:

10(x - 4) - 4(1) = 1(7x + 1)

Now, can you finish it?

 

[TchrQbic]

Can any one show me another way to work this problem:
(x - 4)/2 - 1/5 = (7x + 1)/20

I tried moving (7x + 1)/20 to the left of the equal sign and I got lost. Was I right or am I doing the problem all wrong.

Since this problem involves three fractions, a useful first step is to multiply all terms by the lowest common denominator, which will clear the fractions. For denominators 2, 5, and 20, the LCD is 20.

20(x-4)/2 - 20(1/5) = 20(7x+1)/20

Notice that each denominator will cancel into the 20 multiplier, so that you then have

10(x-4) - 4(1) = 1(7x+1)

Can you finish it from there?

 

[soroban]

Hello, Tresa332006!

Can any one show me another way to work this problem:
\(\displaystyle \;\;\;\frac{x\,-\,4}{2}\,-\,\frac{1}{5}\;=\;\frac{7x\,+\,1}{20}\)

When there are fractions in an equation, multiply through by the LCD, 20.

\(\displaystyle \;\;\;\not{20}^{10}\,\cdot\left(\frac{x\:-\:4}{\not{2}}\right)\,-\,\not{20}^4\,\cdot\left(\frac{1}{\not{5}}\right)\;\:=\;\:\not{20}\,\cdot\left(\frac{7x\,+\,1}{\not{20}}\right)\)

We have: \(\displaystyle \.10(x\,-\,4)\,-\,4\:=\:7x\,+\,1\;\;\Rightarrow\;\;3x\,=\,45\;\;\Rightarrow\;\;x\,=\,15\)

 

[Tresa332006]

Thank you all for the help. I was begining to think that I was doing it wrong until I check the post again and saw that I had the same answer Soroban gave. You guys help me alot thank you.