Table of Contents
Last modified on August 14th, 2024
Equations with fractions, also known as fractional equations, are mathematical expressions where one or more terms involve fractions.
Here are a few examples of fractional equations.
To solve them, we use the balancing method to apply the inverse operation to both sides of the equation. Then isolate the variable on one side of the equation to get its value.
This process involves the following steps to get the solution from an equation:
Let us solve an equation ${\dfrac{5x+7}{2}=-10}$
Finding the Least Common Denominator (LCD)
Here, the denominator is 2
LCD = 2
Multiplying the Least Common Denominator (LCD)
On multiplying both sides of the equation by 2, we get
${2\times \left( \dfrac{5x+7}{2}\right) =-10\times 2}$
⇒ ${5x+7=-20}$
Isolating the Variable on One Side
⇒ ${5x=-20-7}$
Simplifying
⇒ ${5x=-27}$
Dividing Both Sides by 5
⇒ ${x=\dfrac{-27}{5}}$
Thus, the solution is ${x=\dfrac{-27}{5}}$
Problem – Solving equations with fractions when there is VARIABLE IN THE DENOMINATOR
Solve: ${3-\dfrac{4}{x}=1}$
Here, the denominator is x
LCD = x
Now, ${3-\dfrac{4}{x}=1}$
⇒ ${x\left( 3-\dfrac{4}{x}\right) =1\times x}$
⇒ ${3\times x-\dfrac{4}{x}\times x=1\times x}$
⇒ ${3x-4=x}$
⇒ ${3x-x=4}$
⇒ ${2x=4}$
⇒ ${x=2}$
Thus, the solution is ${x=2}$
Problem – Solving equations with SINGLE FRACTION on EACH SIDE
Solve the equation ${\dfrac{4p}{5}+12=0}$
Here, the denominator is 5
LCD = 5
Now, ${\dfrac{4p}{5}+12=0}$
⇒ ${5\left( \dfrac{4p}{5}+12\right) =0\times 5}$
⇒ ${\dfrac{5\times 4p}{5}+5\times 12=0}$
⇒ ${4p+60=0}$
⇒ ${4p=-60}$
⇒ ${p=\dfrac{-60}{4}}$
⇒ ${p=-15}$
Thus, the solution is ${p=-15}$
Problem – Solving equations with MULTIPLE FRACTIONS on EITHER SIDES
Solve the fractional equation ${\dfrac{x}{3}+\dfrac{7x}{2}=5}$
Here, the denominator is 3 and 2
LCD = 6
Now, ${\dfrac{x}{3}+\dfrac{7x}{2}=5}$
⇒ ${6\left( \dfrac{x}{3}+\dfrac{7x}{2}\right) =5\times 6}$
⇒ ${6\times \dfrac{x}{3}+6\times \dfrac{7x}{2}=30}$
⇒ ${2x+3\times 7x=30}$
⇒ ${2x+21x=30}$
⇒ ${23x=30}$
⇒ ${x=\dfrac{30}{23}}$
Thus, the solution is ${x=\dfrac{30}{23}}$
Solve: ${\dfrac{3}{2x}+1=\dfrac{2x-3}{x}}$
Here, the denominator is 2x and x
LCD = 2x
Now, ${\dfrac{3}{2x}+1=\dfrac{2x-3}{x}}$
⇒ ${2x\left( \dfrac{3}{2x}+1\right) =2x\left( \dfrac{2x-3}{x}\right)}$
⇒ ${\dfrac{3}{2x}\times 2x+1\times 2x=\dfrac{2x-3}{x}\times 2x}$
⇒ ${3+2x=2\left( 2x-3\right)}$
⇒ ${3+2x=4x-6}$
⇒ ${4x-2x=3+6}$
⇒ ${2x=9}$
⇒ ${x=\dfrac{9}{2}}$
Thus, ${x=\dfrac{9}{2}}$
Last modified on August 14th, 2024