Summarize this article:
217 LearnersLast updated on October 23, 2025

Solving an equation means determining the value of the variable that makes the equation true. We find the values of the variable that make both sides of the equation equal. A linear equation in one variable has one solution. A system of linear equations with two variables requires at least two equations to find both values.
Solving equations is the process of finding the unknown variable that makes both sides of the equation equal. An equation is a mathematical statement where two expressions, involving a variable, are equal.
In these equations, the LHS and RHS can be interchanged, as both sides represent the same value.
There are different ways to solve an equation depending on its type, such as linear, quadratic, rational, or radical equations.
Solving an equation includes using mathematical operations to isolate the variable to find the value of the unknown variable. This value is found by isolating the variable using mathematical operations. Let’s look at the steps to solve an equation.
We isolate the variable on one side of the equation after completing these steps.
The linear equation in one variable is expressed in the form \(ax + b = 0\), where a and b are real numbers. To solve such equations, follow these steps:
For example:
Solve the equation: \(3(x + 4) = 24 + x\)
Apply the distributive property on the LHS:
\(→ 3x + 12 = 24 + x\)
Group the like terms to one side :
\(→ 3x - x = 24 –12\)
Simplify both sides:
\(→ 2x = 12\)
To isolate x, we divide both sides by 2:
\(→ x = 6\)
Solution: \(x = 6\)
Using the trial-and-error method, we test different values of the variable until we find the one that satisfies the equation.
For example:
Consider the equation \(5x = 35\).
Look for a number that, multiplied by 5, gives 35
We determine \(x = 7\) since \(5 × 7 = 35\).
This method works well for simple equations, but for more complicated ones, it can become challenging and time-consuming.
Some equations can have more than one solution. This is often the case with quadratic equations, which are equations of degree two. The zeroes of a quadratic polynomial are the values that satisfy the equation.
Example:
\((x + 3)(x + 2) = 0\)
This is a quadratic equation that can be solved by writing each factor equal to zero:
\(x + 3 = 0 ⟹ x = -3\)
\(x + 2 = 0 ⟹ x = -2\)
So, the solutions are \(x = -3\) and \(x = -2\).
A quadratic equation is generally written in the form:
\(ax² + bx + c = 0\)
When a quadratic equation is solved, up to two roots are obtained: α and β.
We can solve a quadratic equation in different steps:
Completing the square method systematically solves a quadratic equation by applying the algebraic identity:
\((a + b)^2 = a^2 + 2ab + b^2\)
A quadratic equation can be solved using the factorization method as discussed below:
For example:
Solve: \(2x² + 19x + 30 = 0\)
Find two numbers that add up to 19 and multiply to 60 (2 × 30)
→ 4 and 15
Now rewrite the equation:
\(2x² + 4x + 15x + 30 = 0\)
Group and factor:
\(2x(x + 2) + 15(x + 2) = 0\)
Take the common factor:
\((x + 2)(2x + 15) = 0\)
Now solve each factor:
\(x + 2 = 0 ⇒ x = -2\)
\(2x + 15 = 0 ⇒ x = \frac {-15}{2}\)
When the equation is of the form \(ax² + bx + c = 0\), we use the quadratic formula:
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
To find the solution, we substitute the values of a, b, and c into the formula.
For example:
Solve: \(9x² - 12x + 4 = 0\)
Here, \(a = 9, b = -12, c = 4\)
Apply the quadratic formula:
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \times 9 \times 4}}{2 \times 9} \)
\(x = \frac{12 \pm \sqrt{144 - 144}}{18} \)
\(x = \frac{12 \pm \sqrt{0}}{18} \)
\(x = \frac{12}{18} \)
\(x = \frac{2}{3} \)
Solution: \(x = \frac{2}{3} \)
A rational equation has at least one variable in the denominator. To solve it:
For example:
Solve: \(\frac{2x}{x + 4} = \frac{4}{5} \)
Cross-multiplying gives:
\(5 × 2x = 4(x + 3)\)
\(10x = 4x + 12\)
\(10x - 4x = 12\)
\(6x = 12\)
\(x = 2\)
A radical equation is an equation in which the variable is enclosed in a root. To solve it:
Example:
Solve: \(\sqrt{(2x - 3)} = 5\)
Square both sides:
\((\sqrt{(2x - 3)})² = 5²\)
\(2x - 3 = 25\)
\(2x = 28\)
\(x = 14\)
Here are some of the tips and tricks that will be helpful for the learners to master solving equations:
Solving equations is an important mathematical skill that helps students to develop strong problem-solving skills. However, it is common for students to make small errors that lead to incorrect results. Here are a few common mistakes along with tips to avoid them:
Solving equations is a fundamental concept in mathematics, and we use it in different fields. Let’s now learn about their importance in real life. Here are a few real-life applications of solving equations.
Solve: 3x + 5 = 20
\( x = 5\)
We first subtract 5 from both sides:
\(3x = 15\)
Now, divide both sides by 3:
\( x = 5\)
Solve: x/3 + 1/2 = 1
\(x = \frac {3}{2}\)
For equations with fractions, we multiply the whole equation by 6 (LCM of 3 and 2):
\(6 × (\frac {x}{3} + \frac{1}{2}) = 6× 1\)
\( (\frac {6 ×x}{3} + \frac{6 ×1}{2}) = 6× 1\)
\(2x + 3 = 6\)
Subtract 3:
\(2x = 3\)
Divide by 2:
\(x = \frac{3}{2}\)
Solve: 2x² + 3x - 2 = 0
\(x = \frac {1}{2}, x = -2\)
First, identify \(a = 2, b = 3, c = -2\)
Using the formula for formula method:
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(x = \frac{-3 \pm \sqrt{9 + 16}}{4} \)
\(x = \frac{-3 \pm \sqrt{25}}{4} \)
\(x = \frac{-3 \pm 5}{4} \)
\(x = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad x = \frac{-8}{4} = -2 \)
Solve: √(x + 5) = 6
\(x = 31\)
We first square both sides:
\(\left(\sqrt{x + 5}\,\right)^2 = 6^2 \)
\(x + 5 = 36\)
Now, subtract 5 to isolate x:
\(x = 31\)
Solve: 2(x - 4) = 10
\(x = 9\)
Expanding the brackets:
\(2x - 8 = 10\)
Add 8 on both sides: \(2x = 18\)
Divide by 2 to isolate x:
\(\frac {2x}{2} = \frac {18}{2}\)
\(x = 9\)
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.






