LHS and RHS Examples

A linear equation is present in the forms of one-variable, two, or three-variable. In one variable, a linear equation, the standard equation looks like \[ax + b = 0\] where a and b are the constant numbers, and the x is the variable part we need to find. Similarly, the two-variable equation would look like \[ax + by + c = 0\] where a, b, and c are the constant and x and y are the variables. An example of a linear equation will be \[6 + x = 13\].

How to Solve Linear Equations?

To solve the linear equation with one variable method like LHS=RHS is used. In this method, the variable part and the constant part of the equation are separated from each other so that the value of the variable part can be found. The value is then put into the initial equation to check whether the correct value was found.

To solve the linear equation with two variables, methods like

• Substitution method

• Elimination method

• Cross multiplication method, and

• Plotting the graph method.

How to Find LHS and RHS?

Every algebraic equation has two sides: Right Hand Side (RHS) and Left Hand Side (LHS). In the case of an equation, the two sides are equal, that is, the left and right sides are equal. Let’s understand this using the LHS and RHS example.

\[2x + 5 = 15\] in this equation we have \[2x + 5\] on our LHS and 15 on our RHS. To prove that LHS is equal to RHS we need to find the value of the x.

For that, we will first assume that LHS and RHS are equal and will transfer the constant part of the equation to the RHS and solve it.

\[\begin{array}{l}2x + 5 = 15\\2x = 15 - 5\\x = \dfrac{{10}}{2}\\x = 5\end{array}\]

Now, we will put the value of x as 5 in our initial equation.

So, LHS= \[\begin{array}{l}2x + 5\\2(5) + 5\\10 + 5\\15\end{array}\]

As we got 15 on the LHS and it was given that RHS is also 15. Now we can say that LHS is equal to RHS.

Examples of Solving Linear Equations

Now, let’s work on some examples to understand how to solve linear equations with one variable.

To prove the given equation LHS is equal to RHS \[10x + 7 = 13 - 5x\].

Step one: assume LHS is equal to RHS.

Step two: Place the constant on RHS and the variable on the LHS.

Step three: Solve and find the value of the variable

Step four: Put this value first in the initial equation on LHS and then on RHS. If both sides have the same answer, then LHS would be equal to RHS.

Ans: Assuming LHS is equal to RHS.

\[\begin{array}{l}10x + 7 = 13 - 5x\\10x + 5x = 13 - 7\\15x = 6\\x = \dfrac{6}{{15}}\\x = \dfrac{2}{5}\end{array}\]

Putting the value of x on the LHS.

\[\begin{array}{l}10x + 7\\10(\dfrac{2}{5}) + 7\\4 + 7\\11\end{array}\]

Putting the value of x on the RHS.

\[\begin{array}{l}13 - 5x\\13 - 5(\dfrac{2}{5})\\13 - 2\\11\end{array}\]

As the value of LHS and RHS is 11 so; we can say that LHS is equal to RHS.

Conclusion

Linear equations can be present in one, two, or three-variable forms. The method to solve one variable equation is by finding out the value of the variable. Then putting it in the initial equation to check whether LHS is equal to RHS. If LHS is not equal to RHS, then the value of the variable you found is incorrect, or the equation does not have a value that would equate them.

Sample Questions

1. What is LHS?

a. Left Hour Sign

b. Left Hand Side

c. Less Hour Side

d. Left Home Sweet

Ans: Left Hand Side

2. Does LHS always equal RHS?

a. Yes

b. No

Ans: Yes

Explanation: LHS is always equal to RHS if the value of the variable we found out is right and there is no exception in the equation.

3. When the RHS and LHS are not equal then that equation is called

a. linear equality

b. linear inequality

c. Algebraic equality

d. Algebraic inequality

Ans: Linear Inequality

Explanation: When RHS and LHS are not equal to each other then the equation would have no solution or would have infinite solutions. Such equations are known as linear inequality equations.