Linear Equations in One Variable Made Simple

Things to Remember When Solving Linear Equation Exercises

• To simplify an equation we can multiply or divide the equation on both sides of the equation 

For example, let's take 3x+6= 18 

$3 x+6=18$

$\dfrac{3 x+6}{3}=\dfrac{18}{3}$ (divide the equation by 3 on both sides to simplify)

$x+2=6$

$x=4$

• If the same number is added, subtracted, multiplied, or divided into both sides of a linear equation, the solution remains unchanged.

• When a number is taken from one side of the equal sign to the other, the sign associated with it (-, +, x, ÷) is changed to its opposite sign. 

• If the sign (- or + ) is equal on both sides, they can be canceled out.

Example : -2x = -4 

=  2x = 4         (the minus side on both sides is canceled) 

• Knowing the rules of adding, subtracting, dividing, and multiplying integers is very useful in solving equations.

Solved Examples 

Q 1: Solve the following equations:

$\dfrac{(x-3)}{4}+\dfrac{(x-1)}{5}-\dfrac{(x-2)}{3}=1$

Ans: $\dfrac{(x-3)}{4}+\dfrac{(x-1)}{5}-\dfrac{(x-2)}{3}=1$

So, first of all we are going to add the above equation. For that, we have to take the LCM of the denominator, and here it is 60.

$\Rightarrow \dfrac{\{15(x-3)+12(x-1)-20(x-2)\}}{60}=1$

Now taking the denominator of LHS to the RHS

$\Rightarrow \{15(x-3)+12(x-1)-20(x-2)\}=60$

Now solving the bracket by expanding the multiplication.

$\Rightarrow 15 x-45+12 x-12-20 x+40=60$

Sequencing the like term together to do the further calculation.

$\Rightarrow 15 x+12 x-20 x-45-12+40=60$

After solving, we get

$\Rightarrow 7 x-17=60$

Taking the constant term to the RHS

$\Rightarrow 7 x=60+17$

On further solving, we get

$\Rightarrow 7 x=77$

Solving for the value of x, we get

$\Rightarrow x=\dfrac{77}{7}$

$\Rightarrow x=11$

Q 2: Solve the following equations:

$9-2(x-5)=x+10$

Ans: $9-2(x-5)=x+10$

First of all, we will expand the bracket by doing the multiplication.

$\Rightarrow 9- 2x+10=x+10$

Now adding the constant term together in LHS, we get

$\Rightarrow -2 x+19=x+10$

Keepimg the variable at one side and constant on the another side, so we get

$\Rightarrow -2 x-x=-19+10$

Now on solving, we get

$\Rightarrow -3 x=-9$

Since both side have negative sign, so it will cancel out each other, our equation will become; 

$\Rightarrow 3 x=9$

Solving for the value of x, we get

$\Rightarrow x=3$ 

Q 3:  Solve the following equations:

2x – 3 = 7

Ans: Step 1: Add 3 to both sides. 

$\Rightarrow 2x – 3 + 3 = 7 + 3$

The next step is dividing both sides by 2. 

$\Rightarrow \dfrac{2x}{2} = \dfrac{10}{2}$ 

Therefore x = 5 

Q 4: Solve the following equations:

2x – 3 = x + 2

Ans: 2x = x + 2 + 3 

$\Rightarrow 2x = x + 5$ 

$\Rightarrow 2x – x = x + 5 – x$ (subtracting x from both sides)

Therefore x = 5

Q 5:  Solve the following equations 

72x – 7 = 28x – 5

Ans: Taking variable at one side and constant at other side, we get

72x – 28x = 7 – 5

Solving the LHS and RHS

$\Rightarrow 44x = 2$

Solving for the value of x, we get

$\Rightarrow x = \dfrac{1}{22}$

Therefore x = $\dfrac{1}{22}$

Linear Equation Practice Problems

Q 1: Solve the following equation

3x - 2 = 4

Ans: 2

Q 2: Solve the following equation

$\dfrac{(8 x-5)}{(7 x+1)}=-\dfrac{4}{5}$

Ans: $\dfrac{21}{68}$

Q 3: Solve the following equation

17x - 5 = -$\dfrac{6}{7}$ 

Ans: $\dfrac{29}{119}$ 

Q 4: Solve the following equation

49x + 3 = 10

Ans:  $\dfrac{1}{7}$ 

Summary 

A linear equation, which can be written as the combination of variables (or unknowns) and the coefficients, which are frequently real values and it has equal sign. The coefficients, which can be any expressions as long as they don't contain any of the variables, can be thought of as the equation's parameters.

The coefficients must not all be 0 in order for the equation to have any sense. An alternative method for creating a linear equation is to equalize a linear polynomial over a field, from which the coefficients are drawn, to zero. The numbers that, when used to replace the unknowns in such an equation, result in equality, are the solutions.