Thread: the journey continues... Fractional Equations

Hmmm, I was doing well until I came up against these little buggers.

4x + 3 = 6 and
2x - 1

3x - 2 = -5
x + 4


Solve for x

I come up with x = -9/8 on the first, whilst the text book has 8/9 and second x = 9 and -9/4 respectively.

any help would be greatly appreciated.

P.S. it was hard enough to write out the formulas above, can anyone lead me to where I might learn a quicker way. thanks again.

Chris.

Originally Posted by nativecat

Hmmm, I was doing well until I came up against these little buggers.

4x + 3 = 6 and
2x - 1

3x - 2 = -5
x + 4


Solve for x

I come up with x = -9/8 on the first, whilst the text book has 8/9 and second x = 9 and -9/4 respectively.

any help would be greatly appreciated.

P.S. it was hard enough to write out the formulas above, can anyone lead me to where I might learn a quicker way. thanks again.

Chris.

(1)



x=9/8

(2)



x=-9/4

It is a very good idea, when solving equations with fractions, to multiply through by the least common denominator, getting rid of all fractions as mathaddict did. After solving that equation, you should be sure to check by putting the value into the original equation to see if it satisfies that equation. It doesn't happen here but, it is possible that multiplying both sides of an equation by something will introduce new "solutions" that don't actually satisfy the original equation.

One of the key things to remember is simply to do the same thing on both sides.

I'm assuming if you're up to fractional equations you've done normal equations.

Let's say we have we simply divide each side by 2. This makes yeah?

Similarly if we have and we want to solve it we times both sides by x. , then we move the 50 over the other side and from there we solve it for x = 2.

Since we've got the basics down pat the secret is not to be scared of all the different terms on the end of the bottom. Treat it exactly the same.

Sometimes it's easy to visualize if we cheat a little and see it like this.

<<< The is all one term, treat it exactly like you would have x in the above example

So that means times both sides by it, and we have . This simplifies into and you can follow the above steps.

Like I said, don't get confused by the fact there are a few terms down there, simply multiply by the whole lot.