Adding and Subtracting Algebraic Fractions (Leaving Cert Mathematics): Revision Notes

Adding and Subtracting Algebraic Fractions

Algebraic fractions can only be added or subtracted if their denominators are the same. The following property applies to all algebraic fractions :

$$\frac{x}{a}+\frac{y}{a}=\frac{x+y}{a}$$

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$$\frac{x}{a}-\frac{y}{a}=\frac{x-y}{a}$$

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To add / subtract algebraic fractions we do the following steps :

• Find a common denominator of the denominators of the fractions.
• Rewrite each fraction such that they have the same denominator.
• Apply the above properties to add / subtract the fractions.
• Simplify if necessary.

Method 1

In total we have three fractions, let's move them all to one side, and add to deduce the expression to a single fraction.

$$\frac{3}{2x+1}+\frac{2}{5}-\frac{2}{3x-1}=0$$

Now we determine a common denominator. Recall that we can just take the production, so the common denominator is $(2x+1)(5)(3x-1)$.

Now multiply by whatever is missing in order for all fractions to have this common denominator.

$$\frac{3x-1}{3x-1} \cdot \frac{5}{5} \cdot \frac{3}{2x+1}+\frac{2}{5} \cdot \frac{3x-1}{3x-1} \cdot \frac{2x+1}{2x+1}-\frac{2}{3x-1} \cdot \frac{2x+1}{2x+1} \cdot \frac{5}{5} = 0$$

You should start seeing a pattern here. For any fraction, we simply multiply by the denominators of all the other fractions.

Now we can add (or subtract)

$$\frac{[(3x-1)(5)(3)]+[(2)(3x-1)(2x+1)]-[(2)(2x+1)(5)]}{(2x+1)(3x-1)(5)}$$

Simplify the numerator:

$$\frac{12x^2+27x-27}{(2x+1)(3x-1)(5)}=0$$

We are now left with one fraction, the next logical step would be to eliminate the fraction all together. Since a fraction is just a division, we can get rid of it by multiplying both sides by the denominator. In general:

$$\frac{a}{x}=b \implies a=bx$$

So we multiply both sides by $(2x+1)(3x-1)(5)$.

$$12x^2+27x-27=0$$

We are just left with a standard quadratic, factorise as normal and solve for $x$.

$$x=-3 ,x=\frac{3}{4}$$

You'll notice that adding/subtracting fraction gets more complicated as more fractions are included. Here is a shortcut we can take:

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Method 2

First let's remind ourselves of the original equation.

$$\frac{3}{2x+1}+\frac{2}{5}-\frac{2}{3x-1}=0$$

Notice how $\frac{2}{5}$ is just a constant. Our only concern should be to merge the algebraic fractions together. For now, let's bring the constant to the RHS and have all algebraic fractions on the LHS.

$$\frac{3}{2x+1}-\frac{2}{3x-1}=\frac{2}{5}$$

Now let's only focus on adding the two algebraic fractions. Follow the same procedure; the common denominator is $(2x+1)(3x-1)$.

$$\frac{3}{2x+1} \cdot \frac{3x-1}{3x-1}-\frac{2}{3x-1} \cdot \frac{2x+1}{2x+1}=\frac{2}{5}$$

Merge fractions:

$$\frac{3(3x-1)-2(2x+1)}{(3x-1)(2x+1)}=-\frac{2}{5}$$

Simplify the numerator:

$$\frac{5x-5}{(3x-1)(2x+1)}=-\frac{2}{5}$$

We've arrived at a particular algebraic structure where we have a fraction equal to another fraction. We can do cross multiplication. That is, the denominator on the LHS multiplies with the numerator on the RHS, and the denominator on the RHS multiplies with the numerator on the LHS. In general:

$$\frac{a}{b}=\frac{c}{d} \implies ad=bc$$

Apply the rule :

$$5(5x-5)=-2(3x-1)(2x+1)$$

Now solve for $x$ as normal:

$$12x^2+27x-27=0$$ $$x=-3 ,x=\frac{3}{4}$$