Fractions in Expressions and Equations

Simplify $\frac{3x}{5}+\frac{x}{4}$

Solution:

What we will do here is to find a common denominator for the two terms so they can be added. Firstly, we will have to find the LCD for the denominators of the two fractions, 5 and 4. The lowest common divisor for both numbers will be 20. Now we will find the equivalent fraction for both.

If the denominator for the first fraction should now be 20, that means we probably multiplied the initial denominator, 5 by 4. This would mean we will have to multiply the numerator also by 4 to have an equivalent fraction.

$\frac{4}{4}\left(\frac{3x}{5}\right)=\frac{12x}{20}$

We will do the same for the second fraction. 20 as a denominator also means that we must have multiplied 4 (as the denominator) by 5 to have 20 as the new denominator of the equivalent fraction. This means we will have to multiply the numerator also by 5.

$\frac{5}{5}\left(\frac{x}{4}\right)=\frac{5x}{20}$

We now have our new expression:

$\frac{12x}{20}+\frac{5x}{20}$

This becomes a lot easier to solve since all we have to do is add the numerators and maintain the denominators.

$\frac{12x}{20}+\frac{5x}{20}=\frac{17x}{20}$

Since this cannot be simplified any further, we will leave it at that.

Simplify $\frac{ax-b+x-ab}{a{x}^2-abx}$

Solution:

Since we cannot cancel anything in this current expression, we may want to factorize to see what we can make of the situation. We will first group like terms in the numerator by rearranging so that terms containing x will be close together and terms containing b will also be close.

$\frac{ax-b+\hspace{0.17em}x-ab}{a{x}^2-abx}$

$\frac{ax+x-b-ab}{a{x}^2-abx}$

We will now factorize. The common factor in the first two terms in the numerator is x. That can be factored out. The common factor in the last two terms of the numerator is -b, and that can also be factored out.

$\frac{x(a+1)-b(1+a)}{a{x}^2-abx}$

The whole idea of factorizing here is building a common bracket so one can be taken out. Here we have $(a+1)$ and $(1+a)$. Considering the commutative property of addition, both brackets are the same.

This will leave us with:

$\frac{(x-b)(a+1)}{a{x}^2-abx}$

Now, we will factorize the denominator immediately. As ax appears common in both terms, that is what will be factored out.

$\frac{(x-b)(a+1)}{ax(x-b)}$

We are now left with a situation where we can freely cancel out. $(x-b)$ in the denominator will cancel out $(x-b)$ in the numerator. This is true if $x$ is different from $b$.

$\frac{(a+1)}{ax}$

This is the simplest form we can get from this expression.

If we were given the equation $5x+\frac{1}{2}=12$, we would first multiply the equation (which is technically also each term of the equation) by 2.

Solution:

$2(5x+\hspace{0.17em}\frac{1}{2}=12)$

$2\left(5x\right)+2\left(\frac{1}{2}\right)=2\left(12\right)$

After multiplying by 2, the fraction will cancel out.

$10x+1=24$

We will now rearrange the equation to put like terms on different sides of the equation.

$10x=24-1$

$10x=23$

Divide both sides by 10

$\frac{10x}{10}=\frac{23}{10}$

$x=\hspace{0.17em}2.3$

To check that this is indeed the solution of the equation you need to substitute the value of x back into the original equation:

$5(2.3)+\frac{1}{2}=12$

$11.5+\frac{1}{2}=12$

$12=12$

Solve$\frac{3}{2}x+\frac{1}{2}(x-4)=6$

Solution:

An equation with two fractions with the same denominator will have its terms being multiplied with the denominator, as mentioned earlier.

$2\left(\frac{3}{2}x\right)+2\left(\frac{1}{2}\right(x-4\left)\right)=2\left(6\right)$

$3x+x-4=12$

$4x-4=12$

Like terms will now be grouped from this point.

$4x=12+4$

$4x=16$

Divide both sides by 4

$\frac{4x}{4}=\frac{16}{4}$

$x=4$

To evaluate this, you would need to substitute the value of x back into the original equation.

$\frac{3}{2}\left(4\right)+\frac{1}{2}(4-4)=6$

$(3\times 2)+\frac{1}{2}\left(0\right)=6$

$6+0=6$

$6=6$

Solve$\frac{1}{4}x+\frac{3}{2}(2x-1)=2$

Solution:

Our example is quite different from the usual here. Since we have two fractions with different denominators, we will find the LCM for both and multiply that with the equation. The LCM is 4, so

$4\left(\frac{1}{4}x\right)+4\left(\frac{3}{2}\right(2x-1\left)\right)=4\left(2\right)$

$1x+2\left(3\right(2x-1\left)\right)=4\left(2\right)$

We will now expand what is in the parenthesis:

$1x+2(6x-3)=8$

$1x+12x-6=8$

Group like terms:

$12x+1x=8+6$

$13x=14$

Divide both sides by 13:

$\frac{13x}{13}=\frac{14}{13}$

$x=\frac{14}{13}$

To evaluate this, you would need to substitute the value of x back into the original equation.

$\frac{1}{4}\left(\frac{14}{13}\right)+\frac{3}{2}\left(2\right(\frac{14}{13})-1)=2$

$2=2$