Solving Equations having the Variable on Both Sides

An equation consists of two expressions separated by an equal sign “=”. If an expression is represented using an equal sign, the value of one side should be equal to the value on the other side.

Now, previously, we only solved equations having one variable like $2x + 1 = 3$. This equation has a variable on one side as you can see only 2 has the variable x. But, now, we will solve equations having the variable on both sides. For illustration, we take this equation $4x + 2 = 3 - 9x$. In this equation, we have variables on both sides.

Solving Equations having the Variable on Both Sides

Solving equations having the variable on both sides means that there are two same variables in the equation: one is on the left side of the equation and other is on the right side of the equation; for illustration, we take this equation $3x + 2 = 3 - 2x$. In this equation, we have variables on both sides with some constant. These equations can be solved with many methods like addition and subtraction. Lets’s understand the topic by considering the following example.

Solve the given equation: $5x + 2 = 12$

$5x+2=12$

$5x=12-2$

$x = \dfrac{{10}}{5}$

$x=2$

Here, by solving the following equation, we get the value of the variable i.e. $x=2$.

The second example for variables on both sides:

Solve the given equation: $7x + 49 = 14x$

$7x+49=14x$

$49=14x-7x$

$49=7x$

$x = \dfrac{{49}}{7}$

$x=7$

Here, by solving the following equation, we get the value of the variable i.e. $x=7$.

Solve for x Equations

Solve for x is meant to find the value of x in an equation of one variable that is x. Suppose we have an equation $5x - 10 = 0$ and are asked to determine the value of the variable. If the given variable is x in the equation, it is called solve for x. When we find the value of x and substitute it in the equation, we should get L.H.S = R.H.S.

Now illustrate this with some examples:

Solve for \[x:\dfrac{2}{5} = \dfrac{x}{{10}}\]

Cross multiply the fractions,

\[ \Rightarrow 2 \times 10 = x \times 5\]

Solve the equation for x,

\[\Rightarrow x = \dfrac{{20}}{5}\]

Simplify for x,

\[ \Rightarrow x = 4\]

To verify the x value, put the result 4 back into the given equation,

\[\Rightarrow \dfrac{2}{5} = \dfrac{4}{{10}}\]

Cross multiply the fractions,

\[ \Rightarrow 2 \times 10 = 4 \times 5\]

\[\Rightarrow 20=20\]

L.H.S = R.H.S

Solve for Variables

Now we will learn how to solve an equation for a variable. Its steps are similar to solving for x. In solving for x, we only find the value of x but in solving for a variable, we have to find the value of every variable given in the equations. It generally has two equations having two variables in each equation.

Let’s take a look at an example:

Solve for the variables “x” and “y“: $2x - y = 5, 3x + 2y = 11$

\[ \Rightarrow 2x-y = 5\]

Adding y on both sides, we get,

\[ \Rightarrow 2x-y+y = 5+y\]

\[ \Rightarrow 2x = 5+y\]

\[ \Rightarrow x = \dfrac{({5+y)}}{2}\]

The above equation is known as x in terms of y.

Substitute \[x = \dfrac{{(5+y)}}{2}\] in the second equation \[x = \dfrac{{3(5+y)}}{2}+2y=11\]

\[\Rightarrow \dfrac{{3(5+y)}}{2}+2y=11\]

\[\Rightarrow \dfrac{{(15+3y+4y)}}{2}=11\]

\[\Rightarrow \dfrac{{(15+7y)}}{2}=11\]

\[\Rightarrow 15+7y=22\]

\[\Rightarrow 7y=22-15\]

\[\Rightarrow 7y=7\]

\[\Rightarrow y=1\]

Now, substitute y = 1 in \[x = \dfrac{({5+y)}}{2}\]

\[ \Rightarrow x = \dfrac{({5+y)}}{2}\]

\[ \Rightarrow x = \dfrac{({6)}}{2}=3\]

Here, we get the variables of the given system of equations as x = 3 and y = 1.

Solving Equations having Variables on Both Sides

Consider the equation $5x – 4 = 2x + 2$. To evaluate the variable, we need to get all the variable terms to one side and the constant terms to the other side. Next, we combine like terms and then evaluate the variable by multiplying or dividing the expressions.

Step 1: Add and subtract terms to get the variables on one side and the constants on the other.

Step 2: Multiply or divide to evaluate the variable.

Example: Solve $5x – 4 = 2x + 2$.

Step 1: Get all the variable terms to one side and the constant terms to the other side.

$5x – 4 = 2x + 2$

$5x – 4 – 2x + 4 = 2x + 2 – 2x + 4$ (Subtract 2x and add 4 to both sides)

Step 2: Combine like terms.

$5x – 2x = 2 + 4$

$3x = 6$

Step 3: Divide or multiply to isolate the variable.

$3x = 6$ (Divide by 3)

$x = 2$

Solving an Equation

If an equation is satisfied after the variable has been replaced by the solution or the value of variables, then the number is called a solution of the equation. The real number 3 is a solution of the equation $2x-1 = x+2$ since (2)(3) - 1=3+2. Also, 1 is the solution of the equation $(x-1)(x+2) = 0$. A Solution is a value we put in place of a variable that makes the equation true.

For example: Find the solution of the given equation $2x-1 = x+2$.

Solution:

\[ \Rightarrow 2x-1 = x+2\]

Add $(1-x)$ both sides of the equation.

\[ \Rightarrow 2x-1+1-x = x+2+1-x\]

\[ \Rightarrow x = 2+1\]

\[ \Rightarrow x = 3\]

The solution to the equation is $x=3$

Interesting Facts

Some interesting facts about equations having the variable on both sides are pointed out below.

1. If the coefficients of x are the same on both sides of the equation, then the sides will not equal each other; therefore, no solutions will occur.

2. Two equations having the same solution are called equivalent equations.

3. An equation having a single type of variable is called a linear equation.

4. Sir William Rowan Hamilton first discovered the linear equation.

Solved Problems

1) Solve for x in the given equation \[\dfrac{2-3x}{8} =\dfrac{8-x}{12}-\dfrac{11}{12}\].

Solution:

\[\Rightarrow \dfrac{2-3x}{8} =\dfrac{8-x}{12}-\dfrac{11}{12}\]

\[\Rightarrow \dfrac{2-3x}{8} =\dfrac{8-x-11}{12}\]

\[\Rightarrow \dfrac{2-3x}{8} =\dfrac{x-3}{12}\]

Cross multiplying the expressions, we get,

\[\Rightarrow (2-3x)(12) =(x-3)(8)\]

\[\Rightarrow (24-36x) =(8x-24)\]

Adding $(24+36x)$ on both sides, we get,

\[\Rightarrow (24-36x+24-36x) =(8x-24+24+36x)\]

\[\Rightarrow (48) =(44x)\]

\[\Rightarrow \dfrac{48}{44} =x\]

\[\Rightarrow \dfrac{12}{11} =x\]

2) The sum of the two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers.

Solution:

Let the number be x.

Then, the other number became = 9+x

Now, the sum of the two numbers is = 25

According to question, x + x + 9 = 25

\[\Rightarrow 2x+9=25\]

\[\Rightarrow 2x=25-9\]

\[\Rightarrow 2x=16\]

\[\Rightarrow \dfrac{2x}{2}=\dfrac{16}{2}\] (divide by 2 into both sides)

\[\Rightarrow x=8\]

Therefore, the First number is 8.

The second number becomes = x + 9 = 8 + 9 = 17.

Therefore, the two numbers are 8 and 17.

Key Features

• A linear equation has only one or two variables.

• No variable in a linear equation has a power greater than 1.

• We can calculate the value of the variable from the linear equation.

• If a linear equation consists of two variables, then we can write a variable in terms of another variable.

Practice Questions

1. Solve for x : 3(2x – 4) = 4(2x + 4)

Ans: -13

2. A positive number is 5 times another number. If 21 is added to both numbers, then one of the new numbers becomes twice the other new numbers. Find the original numbers.

Ans: 7 and 35

3. If the value of (3 + 2x) is equal to (3 - 2x), then what is the value of 5 + 3x?

Ans: 5