Presentation Transcript
Solving Systems of Equations Algebraically :Solving Systems of Equations Algebraically Using the Substitution and Elimination Methods
Slide 2:Substitution Method: Substitute one equation into the other… For example: Solve 4x + 3y = 4
y = 2x - 7 Substitute the bottom equation into the top (the bottom is already solved for y). 4x + 3(2x – 7) = 4 4x + 6x - 21 = 4 Distributive property 10x - 21 = 4 Combine like terms. 10x = 25
x = 2.5 Substitute the value of x into
either equation. Y = 2x – 7
Y = 2(2.5) – 7 Y = -2 So… the solution is
(-2.5, -2)
Slide 3:Let’s Look at Another: Solve 2x – 3y = 6
x + y = -12 This time you need to solve one of the equations for one of the variable s–
the bottom equation for either x or y is easiest! Let’s solve the bottom equation for x:
x + y = -12 x = -12 – y (subtract y) Substitute into the first equation:
2(-12 - y) – 3y = 6 Solve:
-24 – 2y – 3y = 6
-24 – 5y = 6
-5y = 30
y = -6 Substitute the value of y
into either equation:
x + y = -12
x – 6 = -12
x = -6 The solution is (-6, -6).
Slide 4:Solving by Elimination: The idea with this method is to add the two
equations together and eliminate a variable. Solve: 4x – 2y = 7
x + 2y = 3 Add the two equations together: 5x = 10 Solve X = 2 Like the previous method, substitute the value of x into either equation. X + 2y = 3
2 + 2y = 3 Solve 2y = 1
y = ½ The solution is (2, ½ ).
Slide 5:Here’s another one (with a twist):
3x + 7y = 15
5x + 2y = -4 This time, if we add them together, nothing
cancels out, so we need to create additive inverses. I am going to make the x terms additive inverses – (15 and -15). 5(3x + 7y = 15)
-3(5x + 2y = -4) 15x + 35y = 75
-15x – 6y = 12 Add the equations
and solve 29y = 87
y = 3 The new equations are: Substitute:
3x + 7y = 15
3x + 7(3) = 15 Solve:
3x + 21 = 15
3x = -6
x = -2 The solution is (-2, 3).