Let’s talk about one of the most common—and honestly, super useful—math concepts you'll come across: linear equations! At first glance, they might just look like a bunch of letters and numbers, but once you get the hang of them, they’re actually a piece of cake. A linear equation is basically any equation where the variable’s highest power is just 1. No crazy exponents, no square roots—just straightforward math. That’s why they pop up so often in exams like the SAT, ACT, PSAT/NMSQT, GED, GRE, GMAT, AP, PERT, Accuplacer, and even the MCAT. They’re foundational!

Now, when you’ve got just one variable, like Ax + B = 0, that’s called a linear equation in one variable. It’s all about figuring out the value of x that makes the equation true. Easy, right? But when things get a little more interesting—say, you’ve got two variables—it looks like Ax + By = C. Don’t worry though, because we’ve got strategies like the substitution method and the elimination method to solve these quickly and confidently.

In this article, we’ll break down everything you need to know about linear equations. We’ll cover their different forms, show you step-by-step how to solve them, and give you plenty of practice examples so you can feel 100% ready when you see them on test day!

What is a Linear Equation?

A mathematical statement with an equal to sign in it is termed as an equation. Such equations of degree 1 are termed as linear equations. This also means that none of the exponent in a linear equation has degree greater than 1. Such an equation represents a straight line and the solution of these equations give values when substituted to the original equation make the equation true.

Also, linear equations can have more than one variable and are named as linear equations in one, two, or more variables. Some of the examples of linear equations in one, two, and three variables are mentioned below:

Linear equations in one variable: 4x + 9 = 0, 45x = 9, (3/2)x + 7 = 0.

Linear equations in two variables: 7x + y = 3, 5a + 4b = 2, 6x – 9y = 12.

Linear equations in three variables: x + y + z = 0, a – 2b = 3c, 3x + 12y = (1/2)z.

Different Form of Linear Equations

The way by which we represent any linear equation is termed as linear equation formula. It can be expressed in three ways namely standard form, slope intercept form and point slope form. Let us check these three ways of expressing the linear equation one by one:

Linear Equation in Standard form

We know that a linear equation is a combination of variables as well as constants. A linear equation in one variable in its standard form is written as:

ax + b = 0, here a cannot be zero and x is the variable.

For a linear equation in two variables, the standard form is written as:

ax + by + c = 0, here a, and b cannot be zero and x, and y are the variables.

For a linear equation in three variables, the standard form is written as:

ax + by + cz + d = 0, here a, b, and c cannot be zero and x, y, and z are the variables.

Linear Equations in Slope Intercept Form

The slope-intercept form of a linear equation is the most commonly used form of linear equation. This can be represented as:

y = mx + c

Here, m is the slope of the line and c is the y-intercept of the line.

x, and y are the x and y coordinates of the line respectively.

For instance:

Consider a line y = 3x + 9.

Here m = 3 is the slope of the line and its y-intercept is 9.

For a line that is parallel to x-axis, the x-coordinate is zero. So, the equation becomes:

y = c.

And, for the line parallel to y-axis the y-coordinate becomes zero and the equation becomes:

mx + c = 0.

Linear Equation in Point Slope Form

In this way of representing a line, the equation is formed by considering the points on the x and y plane. It is represented as:

\(\left(y-y_1\right)=m\left(x-x_1\right)\)

Here \(\left(x_1,y_1\right)\) are the coordinates of the point on the line.

Linear Equation Representation on Graph

The graphical representation of a linear equation in one variable, say ‘x’ is represented by a vertical line parallel to the y-axis, and vice versa. However, a linear equation in two variables forms a straight line.

Let us plot a graph for linear equation in two variables using an example:

Example:

Plot a graph of linear equation in two variables x – 3y = 9.

Following are the steps to plot the graph of x – 3y = 9.

Step 1: Convert the equation in the form y = mx + c. This will give us \(y\ =\frac{1}{3}x-3\).

Step 2: Now we have to replace the value of x and find the subsequent values of y in order to create the coordinates.

Step 3: Starting with putting x = 0 in the above equation, we get y = -3. Similarly, we put values of x = 9, we get y = 0.

Step 4: We can take more values of x and find the subsequent values of y to get a set of coordinates for plotting a line.

Let us construct a table for the obtained values:

The coordinates corresponding to the given values are: (0,3), (3,-2), (9,0), and (12,1).

Step 5: Plot these points on the coordinate plane and join the points to get the desired line.

Linear equations in 1 variable

As the name suggests, any linear equation is said to be a linear equation in one variable if it has only one variable in it. It can be represented as Ax + B = 0, here ‘x’ is the variable, A is the coefficient of ‘x’ and cannot be zero. B is the constant.

The variable ‘x’ in this linear equation has only one solution. The degree of such an equation is always one and it is the easiest way to represent any mathematical statement.

Solving a linear equation in one variable is even easier. We simply have to separate the variables and bring them to one side of the equation and constant to the other side to find the value of the unknown variable.

Example:

Let the linear equation in one variable be \(\frac{\left(2x-10\right)}{2}=3\left(x-1\right)\).

Solution: 

In order to solve this linear equation we have to bring together the variable ‘x’ to one side of the equation.

Let us simplify this equation to get the value of ‘x’ step by step:

\(x-5=3x-3\)

\(3-5=3x-x\)

\(-2=2x\)

\(x=-1\)

So, the solution for this linear equation in one variable is x = -1.

Linear equations in 2 variable

Linear equation in two variables is a type of equation that has two variables but the highest power of both the variables is always one. Such equations are of the form Ax +By + C = 0. Here, ‘x’ and ‘y’ are the variables and A, and B are the coefficients of variables ‘x’ and ‘y’ respectively, and C is the constant. Also, A and B cannot be equal to zero.

In order to get the solution for a linear equation in two variables we need to solve two such equations simultaneously, hence, such equations are also called simultaneous linear equations.

There are multiple ways to solve such pair of linear equations namely, substitution method, elimination method and cross multiplication method about which we will read in the later part of the article.

For Example: 6x + 2y = -9 is a linear equation in two variables.

Linear equations in 3 variable

Such linear equations that have three variables with degree one are called linear equations in three variables. Such equations are represented as Ax + By + Cz + D = 0.

Here, x,y,and z are the three variables and A, B, and C are their coefficients. D in the above equation is the constant. In order to solve a linear equation in three variables we need three such equations.

The most common method of solving such equations is a matrix method.

For Example: x + 3y – 4z= -7 is a linear equation in three variables.

Algebraic Method of Solving Linear Equations

In order to solve linear equations in two variables many methods can be used. The most common methods among those are:

1. Substitution method
2. Elimination method
3. Cross multiplication method

Let us discuss each of these one by one:

Substitution Method

To solve the equation using substitution method we have to follow these steps:

Step 1: Find the value of one variable, let us suppose ‘y’ from one equation in terms of the other variable.

Step 2: Substitute the value of ‘y’ so obtained in the second equation.

In this way we shall get an equation in one variable that is ‘x’.

Step 3: Find the value of ‘x’ using the transposition method.

Step 4: Put the value of ‘x’ so obtained in equation one to get the value of ‘y’.

Let us understand this using an example:

Example: Solve the pair of equations x + 2y = 3, and 7x – 15y = 2.

Solution: The given linear equations in two variables are:

x + 2y = 3…(1)

7x – 15y = 2…(2)

Considering equation (1), let us find the value of ‘x’ in terms of ‘y’, we get:

x = 3 – 2y…(3)

Substitute the value of x from (3) in (2), we get:

7(3-2y) – 15y = 2

21 – 14y – 15 y = 2

 

21 – 29y = 2

-29y = 2 – 21

-29y = -19

y = 19/29…(4)

Put this value of ‘y’ from (4) in (3)

x = 3 – 2(19/29)

x = 3 – 38/29

x = (87 – 38)/29

x = 49/29

So, the value of x = 49/29, and y = 19/29.

Elimination Method

Another method used for solving linear equations in two variables is elimination method. Following are the steps involved:

Step 1: Multiply both the equations with a non-zero constant such that the coefficient of any one of the variables in both the equations becomes the same.

Step 2: Looking at the equation, check if addition or subtraction is to be performed in order to eliminate one variable completely.

Step 3: If the resultant is a true statement, we say that the pair of linear equations has infinitely many solutions.

Step 4: If the resultant is a false statement, we say that the pair of linear equations has no solution.

Step 5: If we get the resultant has an equation in one variable, solve the equation to get the value of either ‘x’ or ‘y’.

Step 6: Substitute the value of the variable so obtained in any one of the two given equations, in order to get the desired result.

Let us understand this using an example:

Example: Solve the pair of equations 2x + 5y = 20 and 3x + 6y = 12.

Solution: The given linear equations in two variables are:

2x + 5y = 20…(1)

3x + 6y = 12…(2)

In order to solve this pair of linear equations using the elimination method, we have to multiply equation (1) by 3 and equation (2) by 2. We get:

6x + 15y = 60…(3)

6x + 12y = 24…(4)

Subtract equation (4) from equation (3), we get:

3y = 36

y = 36/3 = 12

Put the value of y in equation (1), we get:

2x + 5(12) = 20

2x + 60 = 20

2x = -40

x = -20

So, the solution for the given pair of linear equations is x = -20 and y = 12.

Cross Multiplication Method

This is another convenient method to solve linear equations in two variables. Steps involved in solving the equations are as follows:

Step 1: Keep in mind that using cross multiplication method the result are obtained in the form of ratios: \(\frac{x}{?}=-\frac{y}{?}=\frac{1}{?}\).

Step 2: To find out the term below x, strike out the column containing the coefficients of x and cross-multiply and subtract the other two columns.

Step 3: To find out the term below -y, strike out the column containing the coefficients of y and cross-multiply and subtract the other two columns.

Step 4: To find out the term below 1, strike out the column containing the constant term and cross-multiply and subtract the other two columns.

Step 5: Compare the denominator of x and -y with that of 1, we get the value of x and y.

Let us understand this using an example:

Example: Solve the pair of equations 2x + 3y – 11 = 0, and 3x + 2y – 9 = 0.

Solution: To find the solution of the given pair of linear equations, let us first write them in the form of three columns:

2 3 -11

3 2 -9

The solution equality will be of the following form:

 

\(\frac{x}{?}=-\frac{y}{?}=\frac{1}{?}\)

Now we have to find the values for the question mark in the above equation.

Finding the denominator of x:

Striking out the column with coefficients of x and cross multiplying and subtracting the remaining columns:

-27-(-22)=-5 ———— (i)

Finding the denominator of -y:

Striking out the column with coefficients of y and cross multiplying and subtracting the remaining columns:

-18-(-33)=15 ———— (ii)

Finding the denominator of 1:

Striking out the column with constant terms and cross multiplying and subtracting the remaining columns:

4-9=-5 ———— (iii)

The last part of the equality becomes \(\frac{1}{-5}\)

Combining all the equations in our equality:

\(\frac{x}{-5}=-\frac{y}{15}=\frac{1}{-5}\)

⇒\(x=\frac{-5}{-5}=1\)

⇒\(y=\frac{-15}{-5}=3\)

⇒x=1 and y=3

So, the solution for the given pair of linear equations is x = 1 and y = 3.

Linear Equations Solved Examples:

Que 1: The sum of two numbers is 44. If one number is 10 more than the other, find the numbers by framing a linear equation.

Solution: 

Let the two numbers be ‘x’ and ‘y’.

Given that the sum of two numbers is 44, so the equation becomes x + y = 44…(1)

Also, one number is 10 more than another, so the second equation is x – y = 10…(2)

Solving the equations using elimination method:

As we can see that the coefficients of x and y are the same in both equations so we need not multiply the equations with any constant.

Adding equation (1)and (2) to eliminate ‘y’, we get:

2x = 54, or x = 27.

Put this value of x in any one of the given equations, we get:

27 + y = 44,

y = 44 – 27

y = 17

So, the values of x = 27 and y = 17 is the solution of the given pair of linear equations.

Que 2: A boat running downstream covers a distance of 20 km in 2 hrs, and while running upstream it covers the same distance in 5 hrs. What is the speed of the boat in still water?

Solution: Let us suppose the speed of the boat in still water is u km/hr and that the speed of the stream is v km/hr.

Speed downstream = (u + v) km/hr…(1)

Speed upstream =(u – v)km/hr…(2)

Given that: The boat covers a distance of 20 km in 2hr while running downstream and 20 km in 5 hrs while running upstream, so the speed of the boat can be calculated as:

Speed = Distance/ Time

Speed of the boat downstream = (20/2)km/hr = 10 km/hr…(3)

Speed of the boat upstream = (20/5)km/hr = 4 km/hr…(4)

From equation (1), (2), (3), and (4), we get:

u + v = 10, and u – v = 4

Solving the two equations simultaneously, we get:

u = 7 and v = 3

So, the speed of the boat in still water is u km/hr, i.e. 7 km/hr.

Conclusion

Linear equations might seem tricky at first, but once you break them down, they’re actually super straightforward and super useful! Whether it’s one, two, or even three variables, knowing how to solve them gives you a strong edge in exams like the SAT, ACT, GRE, and more. Mastering methods like substitution, elimination, and cross multiplication makes solving linear equations a breeze. Keep practicing—you’ve totally got this math skill locked down!