Algebra question - when you have a fraction like 3/4x where it's 3/4 and then the x is x in the denominator or

Question

numerator? How do you solve something then like

(x-2)/(x) - (7/3)x = (3/4)x

( I saw that question on yahoo answers and didn't see how the person solved it )
thanks

Answer 1

For the first part of your question, fractions don't translate to this format very will. By order of operations, 3/4x would mean divide three by four, and multiply the quotient by x.
Looking at it, though, it is ambiguous. It would be getter to specify the expression as either (3/4) x or 3/(4x). You did a very good job of grouping the terms in the equation you want solved.
Now, as for the equation:

(x-2) / x - (7/3) x = (3/4) x

We have 4, 3, and x as denominators. Let's multiply both sides of the equation by 4*3*x, or 12 x, to get rid of those fractions. *Make a note: x cannot be zero, since x+0 would leave (x-2) / x undefined.

12x (x-2) /x - 12x (7/3) x = 12x (3/4) x
Simplify:

12(x-2) - 4x(7) x = 3x (3) x

12x - 24 - 28x^2 = 9x^2

Subtract 9x^2 from both sides:

12x - 24 - 28x^2 - 9x^2 = 9x^2 - 9x^2

12x - 24 - 37x^2 = 0

Now we'll rearrange the terms in decending order:

-37x^2 + 12x - 24 = 0

We have a second-degree equation, with nothing that is easily factored, so we will use the quadratic formula.

The equation has the form ax^2 + bx + c = 0 where
a= -37, b = 12, and c = -24.

according to the quadratic equation,

x = [-b +/- the square root of (b^2-4ac) ] / (2a)

In this case, (b^2 - 4ac) is a NEGATIVE NUMBER, and your calculator will give you some nasty message when you ask it what the square root of a negative number is. The bottom line is, this equation has no real roots. (Both values for x that satisfy the equation are imaginary numbers.)

Rather than go into further detail here, I would recommend doing some work with the quadratic equation. Also, rather than trying to type this stuff out, get out the old pencil and paper and go to work on it. The quadratic equation, as discussed above, is used to solve second-degree equations. (Equations with one variable, whose highest exponent is 2.)

Sorry about getting so wordy with a math answer, but have fun with it.

Answer 2

When the variable you're solving for is in the denominator of a fraction, it's often a good idea to multiply the whole equation by that denominator, and then go from there.

In your example:
(x-2)/(x) - (7/3)x = (3/4)x

Multiply both sides of the equation by "x", and you get this:
(x-2) - (7/3)x² = (3/4)x²

Notice that the "x" canceled out of the denominator of the first term.

At this point (at least in this example), you now have an equation with "x" and "x²" in it. This is a quadratic equation, and in general you will have to use the quadratic formula to solve it.

Answer 3

you can consider x to be in the numerator, or not part of the fraction at all.

(x-2)/(x) - (7/3)x = (3/4)x (multiply everything by x)
(x-2) - (7/3)x^2 = (3/4)x^2 (move everything to one side)
x - 2 - (37/12)x^2 = 0
Then use the quadratic formula

Answer 4

(x-2)/(x) - (7/3)x = (3/4)x
we have 4 x the 2nd one is only in the denominator (x)
if (7/3)x so x is in the numerator
but if 7/3x so x is in the denominator

Answer 5

to make your life easier, you would want to get rid of the denominators, so you would multiply the ENTIRE problem by 3,4,and x. so you would get:
(3)(4)(x) [ (x-2)/(x) - (7/3)x = (3/4)x ]
.............. [ {(3)(4)(x-2)} - {(4)(x)(7x)}= {(3x)(3x)}
since the first term has an x in the denom. you only multi. by 3 &4 and cancel out the x, since the second term had a 3 in the denom. you only mult. by 4 & x and cancel out the 3 and since the 3rd term has a 4 in the den. you only mult. by 3 & x.
you end up with:
12(x-2) - (4x)(7x)= (3x)(3x)
12x-24-28x^2 = 9x^2
then you just solve the quadratic.
hope that helps :)

Answer 6

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