For solutions to a System of a Equations:
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.
The solutions in the above animation are repeated with a more detailed explanation below.
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 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.
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,
b = the y-intercept, which is the point where the line crosses the y axis
So, for the first equation,
y = 2x + 3,
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.
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,
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)
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!
We have two equations in a system of solutions, y + 2x = 1 & y = 3x - 4
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
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.