SYSTEMS OF EQUATIONS
OBJECTIVE
So, how do we undo Addition?
By subtracting. Whenever you see a + sign and you want to undo it, you subtract whatever is being added.
An example of this:
0 = x + 23
If we want to know what x is, we undo the addition.
So we subtract 23 from the equation.
-23 -23
And we get that x = -23
Some Examples For undoing addition
4. x + 100 = 101
-12 -12
x = 24
-12 -12
x = -5
-5 -5
x = 7
-100 -100
x = 1
So how do we undo Subtraction?
With addition! Whenever you see a - sign and you want to undo it, you add whatever is being subtracted.
An example of this:
x – 20 = 35
If we want to know what x is, we undo the subtraction.
So we add 20 from the equation.
+ 20 + 20
And we get that x = 55
SOME EXAMPLES FOR UNDOING SUBTRACTION
+24 +24
x = 51
+ 50 +50
x = 53
+ 47 +47
x = 49
+57 + 57
x = 157
SO WHAT ABOUT MULTIPLICATION?
We undo multiplication with division. However, sometimes multiplication can be hard to spot.
So an example is this:
If we have something like: 2x = 40
2x is the same as saying 2 * x, so if we want x by itself we divide by 2.
___ ___
2 2
After we divide by 2, we have that x = 20
SOME EXAMPLES FOR UNDOING MULTIPLICATION
___ __
30 30
x = 2
___ ___
400 400
x = 1
____ ___
200 200
x = 1
___ ___
11 11
x = 11
SO THEN HOW DO WE UNDO DIVISION?
Following the same logic, if undoing multiplication is using division, then undoing division must mean we multiply.
If we want to undo division, we need to multiply. The easiest number to multiply by is the denominator so we can get x by itself
5 * * 5
After we multiply the entire equation by 5 we get that x = 125
SOME EXAMPLES FOR UNDOING DIVISION
20 * * 20
x = 100
100 * * 100
x = 2000
30 * * 30
x = 600
40 * * 40
x = 320
2 * * 2
x = 400
SO WHAT ABOUT WHEN YOU HAVE MULTIPLE PROCEDURES GOING ON AT ONCE?
Wait SADMEP?
Yes, to solve for x, we want to work in reverse.
Now, we don’t always have to, but it’s usually the easiest way to solve.
Here’s our example from before:
So, following SADMEP, what we want to do first is undo the subtraction.
To do that, we need to:
And what we’ll get is:
Again, following SADMEP again, what we want to undo now is the division
To do that, we need to:
And finally, we are left with:
So that’s the secret?
That’s the secret.
When we are trying to solve for x, the easier way to work out the problem is to use SADMEP instead.
Now, you don’t have to use SADMEP every time
But when you don’t, things can get really complicated.
Here’s what I mean:
Not using SADMEP
So this time around, we’re not going to solve using SADMEP
Instead, we’re going to use PEMDAS.
So, following PEMDAS, what we want to do first is undo the multiplication.
To do that, we need to:
And what we’ll get is:
Again, following PEMDAS again, what we want to undo now is the subtraction.
To do that, we need to:
And finally, we are left with:
Or, in other words:
So that’s why we use SADMEP
Again, only if we’re trying to find x.
If we’re trying to simplify an equation or expression, we use PEMDAS.
It may seem a little confusing, but let’s try a few more to explain.
EXAMPLE 1:
Solve the following for x:
So, following SADMEP, what we want to do first is undo the subtraction.
To do that, we need to:
And what we’ll get is:
Again, following SADMEP again, what we want to undo now is the division
To do that, we need to:
And finally, we are left with:
So how do we substitute?
Well, honestly, it’s super easy.
Basically, whatever they tell us to substitute, we basically replace the variable with that number.
It sounds way more complicated than it is, so here’s an example:
Example
Solve the following when x = 5:
So, the way that we solve this, is by substituting in 5 for x.
What that means is, where ever we see an x
We replace it with 5.
So:
And now, our answer is y = 13.
So, when x = 5, y = 13.
And that’s it!
SO WHAT IS A SYSTEM OF EQUATIONS?
According to Google: “A system of linear equations is a collection of two or more linear equations involving the same set of variables. ”
Which is essentially right!
So basically a system of equations is a set of equations (usually 2) that share the same variables. In Algebra 1, those variables are x and y.
Solving Systems of Equations
So, how do we solve an equation with 2 variables?
Well, we don’t. To be honest, we have no idea what those two variable can be.
However, when we have 2 equations that share the same variables, that means they share the same x and they share the same y.
So, if they share the same x and the same y, then we should be able to substitute them.
Which is one way we solve systems of equations.
(However, it’s much easier to show then explain)
The Substitution Method
Let’s say we are given:
x + 6 = 4y
4x – 5y = 9
In order to substitute, we need to get one variable by itself.
So in this case, let’s get x by itself in the 1st one.
So:
x + 6 = 4y
- 6 - 6
x = 4y – 6
Now we substitute this into the second equation.
So, wherever we see x, we replace it with 4y – 6
So, the second equation is 4x-5y = 12
We substitute 4y-6 with x and get:
4(4y-6) – 5y = 9
We distribute the 4 and get:
16y – 24 – 5y = 9
Now we combine like terms and get:
11y – 24 = 9
We add 24 and get:
11y = 33
___ __
11 11
And we finally know that y = 3.
Now we plug that into either equation we want to, and we see that:
x + 6 = 4(3)
x + 6 = 12
x = 6.
Now let’s check our answer and see if it works!
So now we can see x = 6 and y = 3, let’s see if this works.
So for our first equation:
x + 6 = 4y
Substituting in, we can see:
6 + 6 = 4(3)
Or
12 = 12
Now let’s try the other equation:
4(x) – 5(y) = 9
4(6) – 5(3) = 9
24 – 15 = 9
9 = 9
So it works!
This is the most common way to solve a system of equations as well.
There are many other ways, however we’re only going to touch on two others from here on out.
But before we do, let’s do one more example, just to check to see if it actually works.
Example 2
Let’s say we are given:
5y = 12x – 32
2y = 6x + 10
So we need to solve for one of the variables.
In this case, let’s try to solve for y in the second equation.
So we have:
2y = 6x + 10
__ __ __
2 2 2
And finally we are left with:
y = 3x + 5.
Now, we substitute in y for the next equation, 5y = 12x – 32
So we have:
5(3x + 5) = 12x – 32
15x + 25 = 12x – 32
+32 +32
15x + 57 = 12x
-15x -15x
57 = -3x
__ ___
-3 -3
-19 = x
Now, we plug in our new number into one of the two equations!
So now we plug x = -19 into the second equation to see what y is.
So for our second equation is:
2y = 6x + 10
Substituting in, we can see:
2y = 6(-19) + 10
Or
2y = -114 + 10
2y = -104
__ ____
2 2
And we get that y = -52
So now we have x = -19 and y = -52.
Again, in order to see if this works, let’s make sure to check our answers by plugging them in!
Now let’s check our answer and see if it works!
So now we can see x = -19 and y = -52, let’s see if this works.
So for our first equation:
5y = 12x – 32
5(-52) = 12(-19) - 32
-260 = -228 – 32
-260 = -260
Now let’s try the other equation:
2y = 6x + 10
2(-52) = 6(-19) + 10
-104 = -114 + 10
-104 = -104
SOLVING BY ELIMINATION
So we know how to solve systems of equations by using substitution, but is there an easier way?
Well, again this is easier to show than to explain, so here’s an example.
The Elimination Method
Let’s say we are given:
6 + 2x = 4y
9 – 2x = y
If we want to use the elimination method, we want to make sure to eliminate a variable.
The way eliminate a variable is we try to either add, or subtract the equations from each other to get rid of a variable.
So in our example, we add:
6 + 2x = 4y
+ 9 – 2x = y
15 = 5y
__ __
5 5
And we get y = 3
Now we pick an equation, then plug it in!
So, let’s try an equation, let’s say:
9 – 2x = y
But this time we replace y with 3
9 – 2x = 3
-9 -9
-2x = -6
-2 -2
And we get that x = 3
So now we need to check if our equation is right.
Now let’s check our answer and see if it works!
So now we can see x = 3 and y = 3, let’s see if this works.
So for our first equation:
6 + 2x = 4y
Substituting in, we can see:
6 + 2(3) = 4(3)
Or
6 + 6 = 12
12 = 12
Now let’s try the other equation:
9 – 2x = y
9 – 2(3) = 3
9 – 6 = 3
3 = 3
So we know this is the right answer!
So we added the equations, what about subtracting?
So when should we subtract instead?
When it eliminates a variable!
So, for example:
14x – 28 = 2y
14x – 42 = 3y
We can see that if we subtract 14x from 14x we’ll eliminate the x variable, so:
14x – 28 = 2y
-(14x – 42 = 3y)
Which is to say:
14x – 28 = 2y
-14x + 42 =-3y
14 = -y
Or
y = -14
Now we just plug in what we found
So:
14x – 28 = 2(-14)
14x – 28 = -28
+28 +28
14x = 0_
x = 0
So, again, now we just check to make sure this is the right answer.
Now let’s check our answer and see if it works!
So now we can see x = 0 and y = -14, let’s see if this works.
So for our first equation:
14x – 28 = 2y
Substituting in, we can see:
14(0) - 28 = 2(-14)
Or
0 -28 = -28
-28 = -28
Now let’s try the other equation:
14x – 42 = 3y
14(0) – 42 = 3(-14)
0 – 42 = -42
-42 = -42
So we know this is the right answer!
So now that we know two ways to solve these, it’s your favorite slide!
Multiplication Elimination
Now that we know we can solve by addition and subtraction, what if we need to multiply one of the equations first to make it easier to solve?
Can we then add them together and then get the correct answer?
Let’s look at an example to see.
EXAMPLE 1
( ) 3
Now let’s use the elimination method
Now we’ve solved for one variable!
So let’s plug it in!
So we know y = -6
x + 4(-6) = -9
x – 24 = -9
+ 24 + 24
x = 15
So now we know that y = -6 and x = 15.
Now let’s make sure to check our answer!
Check our answers!
1
1
-2
5
(5)(1) – 2(-2) = 9
5 + 4 = 9
9 = 9
So our answer is correct!
Example 2
( ) 5
Now let’s use our new equation and eliminate
Now we’ve solved for one variable!
So let’s plug it in!
So we know y = -5
4x + 2(-5) = 10
4x – 10 = 10
+ 10 + 10
4x = 20
4 4
x = 5
So now we know that y = -5 and x = 5.
Now let’s make sure to check our answer!
Check our answers!
1
1
-1
1
(4)(1) – 3(-1) = 7
4 + 3 = 7
7 = 7
So our answer is correct!