This Lesson will cover how we can solve a System of Equations just by using our equations in the system.
Let’s glance over what we covered in the previous lesson. i.e. showing that a point can solve a system if it makes all equations in the system true.
The solutions in the above animation are repeated with a more detailed explanation below.
We have the system, y = 3x + 1 and y + 2x = 11, with the point (2, 7) as a solution.
Firstly, show that the point (2, 7) is a solution to the system.
As we recall, to show a point is a solution to a system, we must show it makes both equations true. This is done by entering that point into both equations.
Remember, a solution to a system is a point where the equations meet. An equation is the description of the points that make up a line and we want the point described by both equations..
Let’s check that the point is a solution to the system (a point on where both lines meet).
Enter (2, 7) into y = 3x + 1
(7) = 3(2) + 1
7 = 6 + 1
7 = 7
It makes the first equation true.
now for y + 2x = 11
(7) + 2(2) = 11
7 + 4 = 11,
11 = 11
Therefore, (2, 7) makes both equations true, meaning it is a point on both lines, and is a solution to the system!
Next, substitute 3x + 1 (from y = 3x + 1) into the other equation in place for y, then rearrange and solve for x.
This will be the new concept we introduce in this lesson. Remember, equations are descriptions of how two points interact with each other and how they change value based on what the other point is.
We can rearrange our equations to give us a representation of what one of the points will be based on the other point. When we have a description of that point, we can think of it as having a value for that point. We can then use that description and sub it into another equation and this will give us the point where the two equations meet.
Here we have a description of the point y based on x, and we can use that description of y, and sub it into the other equation, as we will be left with only one variable, x, we can then move onto solving for x. When we solve for x, this will give us the x value where the lines meet.
Subbing in the description of y which is the expression 3x + 1 into the other equation in our system, y + 2x = 11,
(3x + 1) + 2x = 11
If it helps, think of it like you are subbing in a value for y, like subbing in y = something into the equation.
3x + 1 + 2x = 11
5x + 1 = 11
Subtract 1 from both sides
5x + 1 - 1 = 11 - 1
5x = 10
Divide both sides by 5
(5x) ÷ 5 = (10) ÷ 5
x = 2
This also gives us the x value for where the two lines meet.
Finally, use the x value from above and sub it into both equations (this will determine the y point where the two equations meet and hence the solution to the system) then state the solution to the system and comment on the result from each substitution.
So, we have our x value, x = 2, which is the x value for where the two equations meet, now we need to find the y value for where they meet. Let’s sub in x into each equation.
Let’s start with, x = 2 into y = 3x + 1
y = 3(2) + 1
y = 6 + 1
y = 7
Now let’s sub in x = 2 into y + 2x = 11
y + 2(2) = 11
y + 4 = 11
Rearranging for y, subtract 4 from both sides.
y + 4 - 4 = 11 - 4
The solution to the system is x = 2 and y = 7 or (2, 7)
ii) Comment on the results from i)
First, think about why we have the same value for y when we sub x into each equation.
x = 2 is the x-value where both the equations meet
x = 2 is an x-value on both lines
If we sub that x into either of the equations, it will give us the same y-value as that is the point where both the equations are equal.
Solve the following systems by our substitution technique.
y = 2x + 1 and y = -2x - 7
Again, solving a solution means finding the point that makes both equations true. It is the point where both the equations meet.
Here, we can take one of the expressions for y and substitute it into the other equation.
Let's take y = 2x + 1 and put (-2x - 7) in place of the y in the equation,
(the -2x - 7 comes from y = -2x -7)
(-2x - 7) = 2x + 1
Add 2x to both sides,
-2x - 7 + 2x = 2x + 1 + 2x
-7 = 4x + 1
Subtract 1 from both sides,
-7 - 1 = 4x + 1 - 1
-8 = 4x
Divide both sides by 4,
(-8) ÷ 4 = (4x) ÷ 4
-2 = x
So, x = -2 is the x-value for our solution to the system (the point where the equations meet), now we need to find the y-value.
Remember, the solution to the system is the point on both equations, so in order to find our y-value, we can choose either of the equations to sub our x-value into as both are correct.
Let’s sub x = -2 into y = 2x + 1,
y = 2(-2) + 1
y = -4 + 1
y = -3,
So, our solution to the system is x = -2 and y = -3 or (-2, -3)
6x - 3y = 9 and y = -x + 6
Sub in (-x + 6) into 6x - 3y = 9
6x - 3(-x + 6) = 9
6x -3(-x) -3(6) = 9
6x + 3x - 18 = 9
9x - 18 = 9
Add 18 to both sides
9x - 18 + 18 = 9 + 18
9x = 27
Divide both sides by 9
(9x) ÷ 9 = (27) ÷ 9
x = 3
Now, sub x = 3 into one of our equations, let choose, y = - x + 6
y = -(3) + 6
y = 3
So, our point, x = 3 and y = 3 or (3, 3) is the solution to our system.
We can use this method to solve word problems as well.
A red train and a blue train are travelling towards each other on the same tracks. They start 300 miles away from each other. The red train is travelling at 70 miles per hour (70h), the blue train is travelling at 50 miles per hour (50h). How long until the trains meet?
Try and write to equations about what we know from the question.
We can represent the two trains positions by picking one of the trains to be “position zero” and the others trains position to be based off the train at “position zero”, we must also include a representation of how their position has changed over time (their speed).
We know that when the two trains meet, the positions will be equal.
For the red train we have:
Starting at the beginning, when they are 300 miles apart, set red trains position as 0, accounting for speed as well.
position = 0 plus (red trains speed)
p = 0 + 70h
For the blue train we have:
Starting at the beginning, when they are 300 miles apart, subtract the speed as the blue train is moving in the opposite direction, accounting for speed as well,
position = 300 minus (blue trains speed)
p = 300 - 50h
When both trains have met, the positions will be equal, so we can equate them.
300 - 50h = 0 + 70h
Add 50h to both sides
300 - 50h + 50h = 0 + 70h + 50h
300 = 120h
Divide both sides by 120
300 ÷ 120 = 120h ÷ 120
2.5 = h
So, both trains will meet after 2.5 hours.