# Algebraic Equations

## What is an Algebraic Equation?

The Algebraic Equations are the collection of terms having Constants and Variables separated by using Equality Sign $(=)$ to form Left-Hand Side(L.H.S) and Right-Hand Side(R.H.S) and both the sides of the equality sign are perfectly balanced. The operations between any two terms on either side of the equation may be addition or subtraction or both.

## How to Form an Algebraic Equation?

To understand the formation of an Algebraic Equation let us consider an example. Let, Praveen, who is studying in 8th Standard went to a stationery shop to buy $10$ Ball Pens. While handing him over the Pens the storekeeper asked Praveen for a sum of 100 Rupees for the 10 Pens. Before paying the storekeeper, Praveen wanted to calculate the price of a single Pen and he calculated as follows,

$100 \div 10 = 10\,\,Rupees$

The above calculation can also be expressed mathematically with the help of an Algebraic Equation having constants and variables. To express it in terms of an Algebraic Equation, let us consider the price of one pen is $x$. Therefore, the price of $10$ pens will be $(10 \times x)$. Since the price of $10$ Pens is $100$ Rupees, therefore, $(10 \times x)$ is equal to $100$ Rupees. Mathematically,

$10 \times x = 100$

In the above Algebraic Equation, the L.H.S represents the number of pens $(10)$ multiplied by the price of one pen $(x)$ and the R.H.S represents the total cost incurred $(100\,Rupees)$ in buying $10$ Pens. Therefore, both sides of the equation are perfectly balanced.

Now, if Praveen wants to calculate the price of one pen from the above Algebraic Equation, then he has to find the value of $x$ as it represents the price of a single pen as mentioned above.

Therefore,

$x = \frac{{100}}{{10}} = 10$ Rupees

Again, after buying the $10$ pens, Praveen remembers that he has to buy $5$ Pencils for his younger brother who is studying in $5th$ Standard. So, he asked the storekeeper to give him $5$ pencils. While giving the Pencils, the storekeeper said that Praveen has to pay $130$ Rupees in total for both the Pens and Pencils. Now, Praveen tried to calculate the price of each pencil as given below,

Price of $1$ pen = $10\, Rupees$

Price of $10$ pens = $100\, Rupees$

He paid the Storekeeper $130\, Rupees$ in total for $10$ Pens and $5$ Pencils. Therefore,

Cost of Pencils = Total Amount Paid – Price of Pens

Cost of Pencils = $130 -100$

Cost of Pencils = $30\, Rupees$

Cost of $5$ pencils = $30\, Rupees$

Cost of $1$ pencil = $\frac{{30}}{{5}} = 6\, Rupees$.

Now, the above method for calculating the cost of a Pencil is a little confusing and time-consuming too. If Praveen had bought more items like Erasers, Crayons, Rulers, etc., then it will become very much complex for him to calculate the Price of the Pencils. The same can be done in a very simple way by using Algebraic Equations.

As mentioned above, the price of one Pen was $x$. Let us consider the price of one Pencil as $y$. Therefore, the price of $5$ Pencils is $5y$ and the total amount paid is $130\, Rupees$. Mathematically,

$10x+5y=130$

As calculated above, the cost of one pen, $x=10$

Therefore,

$10\times 10+5y=130$

$100+5y=130$

$5y=130-100$

$5y=30$

$y=\frac{{30}}{{5}}$

$y=6\, Rupees$

The cost of a single Pen and Pencil is $10\, Rupees$ and $6\, Rupees$ respectively.

## Examples of Algebraic Equations

• Let, an Algebraic Equation be $2x+3=9$

In the above Algebraic Equation, the variable is $x$. The above equation means that the L.H.S of the equation $(2x+3)$ will be equal to R.H.S $(9)$ for a value of $x$. The value of $x$, for which the L.H.S is equal to the R.H.S is called the solution of the equation.

Some examples of Algebraic Equations are as follows:

• $x+y=3$
• $-x+y+z=4$
• ${x^3} + {y^2} + 10 = 0$

## Solution of an Algebraic Equation

As discussed above, an Algebraic Expression consists of Constants and Variables. The values of the variable or variables for which the L.H.S. of an Algebraic Equation becomes equal to the R.H.S. of the equation is called the solution or solutions of the Algebraic Equation. In other words, the solution of an Algebraic Equation is the values of the variable or variables present in the equation for which the equation is perfectly balanced.

To understand the solution of an Algebraic Equation let us consider the following example:

• $x+3=10$

For the above Algebraic Equation,

$x + 3 = 10$

$x = 10 - 3$

$x = 7$

Now, for $x=7$, L.H.S. of the equation is

$LHS=x+3=7+3=10=RHS$

For $x=7$, the above equation is perfectly balanced. Therefore, $x=7$ is the solution of the given Algebraic Equation.

## Solved Examples on Algebraic Equations

1. Find the solution of the Algebraic Equation $3x+4=16$?

Solution:
Given, $3x+4=16$
$3x = 16 - 4$
$3x = 12$
$x = \frac{{12}}{3}$
$x = 4$
Therefore, $x = 4$ is the solution of the given equation.

2. Find the solution of the Algebraic Equation $3x+2=x-1$?

Solution: Given,
$3x + 2 = x - 1$
Shifting the variables to the LHS and the constants to the RHS of the equation,
$3x - x = - 1 - 2$
$2x = - 3$
$x = \frac{{ - 3}}{2}$
Therefore, $x = \frac{{ - 3}}{2}$ is the solution of the equation.

3. Find the solution of the Algebraic Equation $\frac{1}{3}x + 2 = 2x + 1$?

Solution: Given,
$\frac{1}{3}x + 2 = 2x + 1$

Shifting the variables to the LHS and the constants to the RHS of the equation,

$\Rightarrow \frac{1}{3}x - 2x = 1 - 2$

Taking $3$ as LCM

$\Rightarrow\frac{{x - 6x}}{3} = - 1$

$\Rightarrow\frac{{ - 5x}}{3} = - 1$

$\Rightarrow\frac{{5x}}{3} = 1$, [Removing $-ve$ sign from both the sides of the equation]

$\Rightarrow5x = 3$

$\Rightarrow x = \frac{3}{5}$

Therefore, $x = \frac{3}{5}$ is the solution of the given Algebraic Equation.