Last updated on July 16th, 2025
Solving an equation means determining the value of the variable that makes the equation true. We find the values that satisfy the condition where both sides of the equation are 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. A quadratic equation usually has two solutions, known as roots. There are several methods to solve equations. In this article, we will learn each of these methods in detail.
Solving equations is the process of finding the unknown variable that makes both sides of the equation equal. An equation can be considered a mathematical statement in which two expressions, including 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 can be determined by using mathematical operations to isolate the variable. 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 = a2 + 2ab + b2
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 = -15/2
When the equation is of the form ax² + bx + c = 0, we use the quadratic formula:
x = [-b ± √(b² - 4ac)] / 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 ± b2 - 4ac2a
x = [-(-12) ± √((-12)² - 4 × 9 × 4)] / (2 × 9)
= [12 ± √(144 –144)] / 18
= [12 ± √0] / 18
= 12 / 18
= 2/3
Solution: x = 2/3
A rational equation has at least one variable in the denominator. To solve it:
For example:
Solve: 2x/x + 4 = 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: √(2x - 3) = 5
Square both sides:
(√(2x - 3))² = 5²
2x - 3 = 25
2x = 28
x = 14
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.
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:
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 = 3/2
For equations with fractions, we multiply the whole equation by 6 (LCM of 3 and 2):
2x + 3 = 6
Subtract 3:
2x = 3
Divide by 2:
x = 3/2
Solve: 2x² + 3x − 2 = 0
x = 1/2, x = -2
First, identify a = 2, b = 3, c = −2
Using the formula for formula method:
x = [-b ± √(b² − 4ac)] / 2a
= [-3 ± √(9 + 16)] / 4
= [-3 ± √25] / 4
= [-3 ± 5] / 4
→ x = (2/4) = 1/2 or x = (-8/4) = -2
Solve: √(x + 5) = 6
x = 31
We first square both sides: (√(x + 5))² = 6²
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:
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.