College beginning Algebra: 3x/2 - 5 = 4x + 5/8

Rodneygc

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I have a math problem that i can't figure out I know the answer because it is in the back of the book however I can't seem to figure out the answer. If possible can someone please show me the steps. Thank You.
The problem is: 3x/2-5=4x+5/8
 
Re: College beginning Algebra

Rodneygc said:
I have a math problem that i can't figure out I know the answer because it is in the back of the book however I can't seem to figure out the answer. If possible can someone please show me the steps. Thank You.
The problem is: 3x/2-5=4x+5/8
Get rid of fractions, unless you like them. Multiply everything by 8

8*(3x/2)-8*5=8*(4x)+8*(5/8)
12x - 40 = 32x + 5

1) Why did I pick "8"?
2) Can you take it from there?
 
IF answer per book = -5/12, then you need BRACKETS in your equation...
 
Re: College beginning Algebra

Rodneygc said:
The problem is: 3x/2-5=4x+5/8
What you have posted means the following:

. . . . .\(\displaystyle \L \frac{3x}{2}\,-\,5\,=\,4x\,+\,\frac{5}{8}\)

Is this what you meant? Or did you mean something more along the lines of the following?

. . . . .\(\displaystyle \L \frac{3x}{2}\,-\,5\,=\,\frac{4x\,+\,5}{8}\)

Or something else?

Were the instructions to "solve the equation"? What have you tried?

Thank you.

Eliz.
 
Hello Rodney!

Solve: \(\displaystyle \L \frac{3x}{2} - 5=4x + \frac{5}{8}\)

Mulitply by \(\displaystyle 2\): \(\displaystyle \; {3x} - 10 = 8x + \frac{5}{4}\)

Multiply by \(\displaystyle 4\): \(\displaystyle \; 12x-40=32x+5\)

Subtracting \(\displaystyle 12x\): \(\displaystyle \; - 40 = 20x + 5\)

Subtract \(\displaystyle 5\): \(\displaystyle \; -45=20x\)

Divide by \(\displaystyle 20\): \(\displaystyle \; -2\frac{1}{4}=x\)

Check: \(\displaystyle \; \frac{3(-2\frac{1}{4})}{2} - 5=4(-2\frac{1}{4}) + \frac{5}{8}\)

Subsitution: \(\displaystyle \; -8\frac{3}{8}= -8\frac{3}{8}\)

If that is not what you mean then at least you will have something to help you.
 
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