We will be solving equations involving fractions:

1
We will be solving equations involving fractions
1.
2.
3.
(click for the solutions)
Lets look at the first equation
.In order to solve this equation we must add
When we add fractions together we have to have a
common denominator.
together.
We are going to begin by finding a number that 2
and 5 go into. That number is 10.
________ 10
2 goes into 10 five times, so we need to multiply
the x above 2 by 5.
5 goes into 10 two times, so we multiply x above
5 by 2.
Simplifying gives us
After cross-multiplying we get
To solve, we divide both sides by 28 and simplify
(for the next solution click here)
2
We are going to solve the second equation in
exactly the same way
But this time x is in the denominator.
We have to find a common denominator that x and
3x both go into,
and that is 3x.
_____ 3x
x goes into 3 x three times, so we multiply 5 by
3.
3x goes into 3x ones, so we will multiply 2 by 1.
Simplifying gives us
After cross-multiplying we obtain
To solve, we divide both sides by 6
(for the next solution click ones)
3
In the third equation is
This time the common denominator is
_______________
(x1)
(x2)
2
We multiply 3 by (x2)
and 4 by (x1)
We need to multiply the top
After simplifying we get
We cross-multiply and get
To be able to solve we have to multiply the
brackets, re-arrange an simplify.
We now have a quadratic equation, which we have
to solve.
For the solution click again
4
We will factorise it using a standard method for
factorising quadratics.
2 times (-6) gives us (-12),
We are looking for two numbers that multiply to
give (-12) and add to give (-1).
They are -4 and 3.
So I have
If we factorise the first two terms
We can take out a factor of 2x. This leaves us
with (x - 2)
If we factorise the last two terms
We can take out a factor of 3. This leaves us
with (x - 2)
We now have
We now have a common factor of (x -2 )
By equating both brackets to zero we get two
solutions