How to solve Straight Line Equations

by | May 18, 2022

In the National 5 Maths course, students are expected to be able to solve straight line equations. There are two equations that can be used to solve straight-line equations, which are shown below.

Normally the equation of a straight line is y = mx + c.

What is a Straight-Line Equation?

A straight-line equation can be used to work out the gradient and intercept of a linear equation. The general equation of the straight line is y = mx + c, where m = gradient and c is the y-intercept. The gradient can be calculated using any 2 points selected on the line of the graph. There is another form of the straight-line equation, which is y-b = m(x-a). Both equations will give the same answer, there is just a different approach taken for each to solve the problem.

Equations 1 & 2 show the different equations used.

Eq 1: y = mx + c            Eq2: y – b = m(x – a)

 

straight line equation

Source of image: https://thirdspacelearning.com/gcse-maths/algebra/y-mx-c/

 

How to work out the gradient of a straight line

The gradient is the slope of the straight line. It can be calculated when there are two known points of the line, stating the x and y coordinates. Using the two points on the line, the following equation can be used to calculate the gradient (Eq 3). Where y2 and y1 are the y-coordinates, and x1 and x2 are the x-coordinates of the points on the line.

y2 – y1
________

x2 – x1

Eq3: Calculation of gradient

Example: A line passes through the points (1,3) and (3,7). Find the equation of the line.

Step 1: Find the gradient of the line by using the two points given.

Eq 3: m = 7 – 3/3 – 1          m = 4/2 = 2

Using Eq 3, the gradient can be determined, which is the difference in y-coordinates divided by the difference x-coordinates. Now that we have the gradient, we can move on to step 2 and work out the y-intercept of the equation.

Step 2: Substitute one of the points on the line, into either Eq 1 or Eq 2, and solve for c (the y-intercept). For example, using Eq 1 (y = mx + c), a point (x,y) on the line can be substituted into the equation, which can then be used to solve for the value of c (the intercept).

Sub in point (1,3) into y = mx + c:

y = mx + c
3 = (2×1) + c
3 = 2 + c
c = 1
y = 2x +1

Sub in point (1,3) into y-b = m(x-a):

y – b = m(x – a)
y – 3 = 2(x – 1)
y – 3 = 2x – 2
y = 2x + 1

As you can see from the example above both equations give you the same final answer. Follow the video link below to see a full explanation of how to solve the straight-line equation using the two different straight-line equations.

 

 

If you would like to get in touch with us about Tutoring in Scotland get in touch. 

We have a number of other articles to help with maths please have a look:
How to solve Mole Calculations using Molar Volume
Easy to use guide to passing the National 5 math test in Scotland

Need help studying?

We have friendly tutors to give you support with your exams in biology, chemistry, maths and physics.

Click here to find tutors!

Looking for a tutor to help your child?

At Central Tutors, we offer a range of professional and experienced tutors to help your child get the grades they want!

Click here to find tutors!