Solving Systems of Equations by Graphing

For solutions to a System of an Equations:

  1. For a system of equations made of variables x and y, a point is a solution if it makes all equations in that system true. On a graph, the point is a solution, if it is a point where the equations in the system meet.
  2. The Solution Set of a System of Equations is all the points (x, y) that are a solution to the system. On a graph, the solution set is all the points where the lines for the equations in the system intersect.

X-Y Graph

A reminder of what our x, y graph looks like is below. These graphs are two-dimensional meaning they plot the coordinates of a plane. You can think of it as a flat surface such as your desk or book, you can describe any point on a two-dimensional plane by giving the two coordinates.

example of a x-y graph with 4 quadrants and cartesian grid

Question 1

In each question, we are given a system of equations made of two equations, determine if the point (1, 3) is a solution to this system of equations.

i. System of Equations: y= 4x - 1 & y + 3x = 6

Now we check if the point (1, 3) is a solution to the system of equations above,

So, we have the point (1, 3), so x = 1 and y = 3,

To check if it’s a solution, we plug these two values of x and y into each equation, and check it remains true,

For the first equation,

y = 4x - 1, sub in x = 1 and y = 3,

3 = 4(1) - 1

3 = 3,

Therefore, the point is true for the first equation, let us check the second,

y + 3x = 6, x = 1 and y = 3,

3 + 3(1) = 6

3 + 3 = 6

6 = 6,

Therefore, the point is true for both equations in the system and is therefore a solution to the system.

ii. System of Equations: y = 2x + 1 & y + x = 5

Check the first equation for the point (1, 3)

y = 2x + 1, x = 1 y = 3

3 = 2(1) + 1

3 = 2 + 1

3 = 3,

Therefore (1, 3) makes the first equation true,

For the second,

y + x = 5, x = 1 y = 3,

3 + 1 = 5

4 = 5,

The point (1, 3) does not make the second equation in our system of equations true, therefore, as it does not make all the equations true, it is not a solution to our system of equations.

Question 2

We have the system of equations: y = 2x + 3 & y = 6 - x

Graph both equations on a graph,

Let us first calculate our gradient and our y-intercept for each equation,

Remember, for a line y = mx + b,

M = the gradient = {change in the y value} / {change in the x value}

b = the y-intercept, which is the point where the line crosses the y axis

So, for the first equation,

y = 2x + 3,

The gradient,

m = 2/1

i.e. for every 1 x value we move on the x axis, we move 2 on the y axis.

Our y-intercept, b, = 3,

So, start by plotting the point on the y axis at y = 3,

Now, from that y-intercept point of y = 3 (and x = 0) plot every point, following the gradient, m = 2 / 1,

That is, every 2 y points we move down, we move to the left 1 x point,

And every 2 y points we move up, we move to the right 1 x point,

x-y grid with equation y = 2x + 3 plotted.

Now, for Graphing y = 6 - x,

It might help to rearrange the equation,

y = -x + 6

y = mx + b

As we can see,

m = -1, but it is better to write the gradient as,

m = -1 / 1

b = 6,

First, plot our y-intercept, b = 6, (which is y=6, x=0)

Now, from the y-intercept, use the gradient to plot every point for our equation,

The gradient is m = -1 / 1,

This means,

Every 1 x value you move to the right, move down 1 y value,

Every 1 x value you move to the left, move up 1 y value,

[Note: Remind ourselves that when we have a negative value in the gradient, we plot our points in the “opposite” way as normal.

(when it’s a positive gradient, if we move up one point on the y axis we move to the right on the x axis, and if we move down one on the y axis we move to the left on the x axis)

(when it’s a negative gradient, if we move up one point on the y axis we move to the left on the x axis, and if we move down one on the y axis we move to the right on the x axis)]

x-y grid with 2 equations plotted; y = 2x + 3 and y = 6 - x

Where do the two graphed lines intersect?

As we can see on the graph,

The point where the two lines intersect is x = 1 y = 5, or (1, 5)

Show that the point where the two lines intersect is a solution to the system of equations

Remember, to show a point is a solution to a system of equations, we show that when we put in the x and y values into each equation, it makes both of those equations true.

So, we sub (1, 5) (x = 1 and y = 5) into y = 2x + 3 and y = 6 - x

For y = 2x + 3,

5 = 2(1) + 3

5 = 2 + 3

5 = 5

So, it’s true for this equation

For y = 6 - x,

5 = 6 - 1

5 = 5

It is also true for this equation.

As the point (1, 5) makes both equations in the system true, it is therefore a solution to the system.

What we are doing is taking a point and checking if it's on each of the lines by putting that point into each equation. If that one point is on both lines, then logically it must be a point where the two lines meet!

Question 3

We have two equations in a system of solutions, y + 2x = 1 & y = 3x - 4

  1. Plot each line on a graph
  2. Determine the point at which they intersect
  3. Show that that intersection point is a solution to the system of equations

Starting with y + 2x = 1

We can rearrange it by subtracting 2x from both sides, to look like,

y = -2x + 1,

We can see that our gradient is,

m = -2 / 1

And our y-intercept is,

b = 1

Plot our y-intercept first,

From our y-intercept, plot every point,

Moving down 2 y values and right 1 x value,

And moving up 2 y values and left 1 x value

Drawing our line between those points.

For y = 3x - 4

Our Gradient is,

m = 3 / 1

And our y-intercept is,

b = -4,

Plot our y-intercept,

From our y-intercept use the gradient to plot our points,

For every 3 points we move up on the y-axis, move to the right 1 point on the x-axis

For every 3 points we move down on the y-axis, move to the left 1 point on the x-axis

x-y grid with 2 equations plotted; y + 2x = 1 and y = 3x - 4