Notes on
Systems of Equations
(In the Algebra Units)
Systems of Equations
Systems of equations are two or more equations that are solved together at the same time.
For example: x + y = 13 are a systems of equations� x - y = 1
You have to figure out what values for x and y work in BOTH equations.
For example, x=7 and y=6, works in both of the equations above.
The above two equations could also be equations describing two lines. You would be solving to find the point where the lines associated with both equations cross. In the above example, the lines would intersect at (7,6) on the coordinate grid.
There are a few methods for solving systems of equations:
For both, the main goal is to eliminate one of the variables so that you can solve for the other.
For example, let’s see how to solve this systems of equations
x + y = 23
x - y = 9
Elimination (Addition or Subtraction) Method
For this one, you just add or subtract the two equations so that one of the variables (x or y) is eliminated.
x + y = 23
2x = 32 See how adding the +y and the -y = 0 so � that the y variable is eliminated?�Solve for x x = 16
Then substitute in the 16 for the x in one of the equations and solve for y�For x + y = 23 → 16 + y = 23 → so y = 7
The final answer is (16, 7)
Another example of the Addition or Subtraction Method:
2a + 3b = 28
4a + 0 = 32
a = 8
So, then solve for b
2(8) + 3b = 28� 3b = 12� b = 4
Answer = (8,4)
2a + 3b = 28
0 + 6b = 24
b = 4
So, then solve for a
2a + 3(4) = 28� 2a= 16� a = 8
Answer = (8,4)
By Adding
Or by subtracting
Try these ones (answers are on the next slide)
2x + y = 8
x - y = 4
2x + 3y = 16
-2x + 6y = 2
Answers
2x + y = 8
+ x - y = 4
3x = 12
x = 4
If I plug that in the second equation:
4 - y = 4
y = 0
Answer: (4,0)
2x + 3y = 16
+ -2x + 6y = 2
9y = 18
y= 2
If I plug that in the first equation:
2x + 3(2) = 16
2x + 6 = 16
2x= 10
x = 5
Answer: (5, 2)
Sometimes you have to do an additional step first
If the two systems of equations don’t add or subtract in a way that results in a variable being eliminated, then you may have to multiply one of the equations by a number first, so that when you do add or subtract them, you DO eliminate a variable:
2x + 3y = 9� x - 2y = 1 → You will have to multiply this WHOLE equation by 2 so that� when you subtract them, the x is eliminated
2x + 3y = 9� - 2x - 4y = 2� 0 + 7y = 7
So y = 1 then substitute that y in one of the equations x -2(1) = 1� x = 3 � Solution: (3,1)
Let’s Practice That
5x + y = 8� 2x - 3y = -7 what can you multiply the top equation by so that when we add � the two equations together, it eliminates the y?
Try it on your own, then go to the next slide to see one way � to work out the problem
Let’s Practice That -- the Solution
5x + y = 8� 2x - 3y = -7 if we multiply the top equation by 3, then we can add them � together to eliminate the y.
15x + 3y = 24�+ 2x - 3y = -7 � 17x + 0 = 17� x = 1
Then solve for the y by substituting in the 1 for the y in one of the equations: � 5(1) + y = 8� y = 3�Solution = (1, 3)
Here are a few more:
x + y = 4�2x + 3y = 10
x - y = -4�2x - 3y = -14
x + y = 4�3x - 2y = 7
Here are a few more with answers
x + y = 4�2x + 3y = 10
2x + 2y = 8 �-2x + 3y = 10� -y = -2
y = 2
x + 2 = 4� x = 3� (2, 2)
x - y = -4�2x - 3y = -14
(2, 6)
x + y = 4�3x - 2y = 7
(3, 1)
If you are comfortable with the previous slides, now you can try a problem where you have to multiply BOTH equations by something so that one of the variables is eliminates
This might make more sense when you see the example —>
Sometimes, you might have to multiply BOTH equations
2x + 3y = 14 Here, you will need to multiply the top by 2� 3x - 2y= 8 and the bottom by 3 so that the y’s will be eliminated� (You could also multiply the top equation by 3 and the � bottom by 2 and then subtract so the x’s are eliminated.)� 4x + 6y = 28� 9x - 6y = 24
13x = 52� x = 4
Then 2(4) + 3y = 14� 3y = 6� y = 2 Solution: (4, 2) �
If you are comfortable with subtraction/addition method, now you can try the other method: Substitution
Another method you can use is SUBSTITUTION
x + 2y = 7� 3x + y = 11
Pick one equation and isolate one of variables so it is all by itself on one side. For example: x + 2y = 7 can be rewritten as x = 7 - 2y
So now you know what x is: It is 7 - 2y
NOW, look at the other equation, and wherever you see “x” SUBSTITUTE in the “7-2y” so instead of 3x + y = 11 you would put 3(7 - 2y) + y = 11
Then solve for y → 21 - 6y + y = 11 so → -5y = -10 so → y = 2
Then pick one equation and solve for the other variable x + 2x = 7� Is now x + 2(2) = 7 so x = 3 The solution is (3, 2)
Substitution is probably most helpful when it’s easy to isolate one of the variables:
2x + 3y = 9� x - 2y = 1 → we could rewrite this one as x = 2y+1
Then plug that in the top equation so that it is 2(2y+1) + 3y = 9
Simplify and solve for y → 4y + 2 + 3y = 9� 7y + 2 = 9� 7y = 7� y = 1�Then plug that into one of the original equations → x - 2(1) = 1� x = 3�The solution is (3, 1)
Systems of Equations and Word Problems
The sum of two numbers is 18. Twice the first number plus three times the second number equals 40. Find the numbers.
x + y = 18 → let’s multiply that by 2 → 2x + 2y = 36
2x+3y = 40 → subtract this equation → - 2x + 3y = 40� -y = -4� y = 4
Then, using the first equation: x + 4 = 18� x = 14 � Solution is 14 and 4
Another Word Problem:
Three times the length of the field increased by four times the width equal 24 kilometers. Two times the length decreased by three times the width is -1 kilometer. What are the length and width of the field?
3L + 4w = 24 → let’s multiply that by 3 → 9L + 12w = 72
2L - 3w = -1 → let’s multiply that by 4 → + 8L - 12w = -4� 17L = 68� L= 4
Then, using the first equation: 3(4) + 4w = 24 � 4w = 12 � w = 3�Solution: length = 4 km and width = 3km