Linear Equations Formula

A linear equation is known as an algebraic equation that represents a straight line. It is composed of variables and constants.

There are various ways to represent the linear equations, such as,

Standard Form

The most generalized form of a linear equation in two variables is:

Ax + By = C where A, B, and C are constants, and x and y are variables.

This form is particularly useful for quickly determining the intercepts and is common in algebraic problems.

Example: Convert the equation y = 2/3x - 4 into standard form.

Solution:

We start with the given equation: y = 2/3x − 4
Multiply everything by 3 to eliminate the fraction: 3y = 2x − 12

Rearrange to get it in Ax + By = C form: 2x − 3y = 12

Slope-Intercept Form

This is another highly popular form for representing linear equations, especially when graphing:

y = mx + by where m is the slope of the line, and b is the y-intercept.

This form directly shows how y changes with x and where the line crosses the y-axis.

Example: Find the equation of a line with slope m = −5m and y-intercept b = 7.

Solution:

Since we are given the slope and y-intercept, we directly substitute into the equation:
y = −5x + 7

Point-Slope Form

Useful for when you know a point on the line (x₁, y₁) and the slope

m: y − y₁ = m(x − x₁)

This form is handy for writing the equation of a line when you are given a point on the line and the slope.

Example: Find the equation of the line passing through (3, -2) with a slope of 4.

Solution:

Using the point-slope formula: y − (−2) = 4(x − 3)

Simplify: y + 2 = 4x − 12

Subtract 2 from both sides:

y = 4x − 14 (which is also in slope-intercept form)

Intercept Form

If the line intercepts the axes at (a, 0) and (0, b), then the equation can be expressed as:

\frac{x}{a} + \frac{y}{b} = 1

This is useful when you know where the line crosses the x-axis and y-axis.

Example: Find the equation of the line that intercepts the x-axis at (6,0) and the y-axis at (0, 4).

Solution:

Using the formula: \frac{x}{a} + \frac{y}{b} = 1

Substituting a = 6 and b = 4:

\frac{x}{6} + \frac{y}{4} = 1

Solved Examples

Question 1: Solve for y, 6y – 3 = 0

Solution:

Solving for the value of y,

Adding 3 to both sides of the equation,

⇒ 6y - 3 + 3 = 3
⇒ 6y = 3

Dividing both sides of the equation by 6
⇒ y = 3/6

Simplifying the equation,
⇒ y = 1/2

Question 2: Solve the equation in x, 4/5x -5 = 15

Solution:

4/5x - 5 = 15

Taking constants to RHS,

4/5x = 15 + 5
4/5x = 20
x = 100/4
x = 25

Question 3: There are two numbers, one equal to 7/6 and the other equal to 1/3 times some number x. The sum of these two numbers is 1. Find x.

Solution:

The sum of both the numbers is 1 so the equation will be, 7/6 + 1/3x = 1

Taking all the constants to the R.H.S of the equation.

1/3x = 1 - 7/6
1/3x = -1/6

Multiplying both the side of the equation by 3
3 (1/3x) = 3 × (-1/6)
x = -1/3

Question 4: Solve the equation in x, 3x + 5y = 33, where y = 3

Solution:

We have been provided with an equation 3x + 5y = 33

We have to find the value of x as the value of y is provided in the question y= 3

So, putting the value of y in the equation
3x + 5(3) = 33
3x + 15 = 33

By taking all the constants to the R.H.S of the equation.
3x = 33 - 15
3x = 18
x= 18/3
x = 6

So here the value of x is 6

Question 5: There are two numbers, one equal to 2/4 some number y, and the other equal to 1/3 times some number x. The sum of these two numbers is 3. Find y. And the value of x is x = 2

Solution:

The sum of both the numbers is 3 so the equation will be, 2/4y + 1/3x= 3

Pitting the value of x the equation will be

2/4y + 1/3(2)= 3
2/4y + 2/3= 3

Taking all the constants to the R.H.S of the equation.

2/4y = 3-2/3
2/4y = 4/3
y= 4*4/3*2
y = 16/6
y = 8/3

So, the value of y will be 8/3

Practice Problems

Question 1: Solve for x: 5x + 7 = 12.

Question 2: Solve for y: 9y − 4 = 23.

Question 3: Solve for z: 3z + 11 = 2z + 7.

Question 4: Solve for x: 7x − 5 = 2x + 20.

Question 5: Solve for y: 12y + 4 = 8y − 16.