Download 1 Substitution and Elimination

Solving Linear Systems
Solving Linear Systems
There are two methods of solving a
system of equations algebraically:
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Elimination
Substitution
Solving Systems of Equations
using Elimination
Steps:
1. Place both equations in Standard Form Ax+By=C
2. Determine which variable to eliminate with Addition
or Subtraction.
3. Solve for the variable left.
4. Go back and use the found variable in step 3 to find
second variable.
5. Check the solution in both equations of the system.
Elimination
● The key to solving a system by elimination is getting rid of one variable.
● Let’s review the Additive Inverse property.
● What is the Additive Inverse of: 3x? -5y? 8p? q?
-3x
5y -8p
-q
● What happens if we add two additive inverses?
We get 0. The terms cancel.
● We will try to eliminate one variable by adding, subtracting, or
multiplying the variable(s) until the two terms are additive inverses.
● We will then add the two equations, giving us one equation with one
variable.
● Solve for that variable.
● Then plug back into one of the original equations to find the other
variable.
Elimination – Example #1
Solve the system:
m+n=6
m -n=5
Notice that the n terms in both equations are additive inverses. So if we
add the equations the n terms will cancel.
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So let’s add & solve:
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Plug in to find n:
m+n=6
+ m- n=5
2m + 0 = 11
2m = 11
m = 11/2 or 5.5
5.5 + n = 6
n = .5
The solution is (5.5, .5).
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Elimination – Example #2
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Solve the system: 3s - 2t = 10
4s + t = 6
We could multiply the second equation by 2 and the
t terms would be inverses. OR
We could multiply the first equation by 4 and the second equation
by -3 to make the s terms inverses.
Let’s multiply the second equation by 2 to eliminate t. (It’s easier.)
3s - 2t = 10
3s – 2t = 10
2(4s + t) = (6)2
8s + 2t = 12
Add and solve:
11s + 0t = 22
11s = 22
s=2
Plug in to find the value of t
3(2) - 2t = 10
t=-2
The solution is (2, -2).
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Elimination
Solve the system by elimination:
1.
-4x+ y=-12
4x+2y=6
2.
5x+2y=12
-6x-2y=-14
3.
5x+ 4y = 12
7x - 6y = 40
4.
5m + 2n = -8
4m +3n=2
Substitution
Substitution
To solve a system of equations by substitution…
1. Solve one equation for one of the variables.
2. Substitute the value of the variable into the other
equation.
3. Simplify and solve the equation.
4. Substitute back into either equation to find the value of
the other variable.
Substitution
● Solve the system: x - 2y = -5
y=x+2
Notice: One equation is already solved for one variable.
Plug in (x + 2) for y in the first equation.
x - 2y = -5
x - 2(x + 2) = -5
● We now have one equation with one variable. Simplify and solve.
x - 2x – 4 = -5
-x - 4 = -5
-x = -1
x=1
● Substitute 1 for x in either equation to find y.
y=x+2
1+2=3
● The solution is (1, 3).
Substitution
● Let’s check the solution. The answer (1, 3) must check
in both equations.
x - 2y = -5
y=x+2
1 - 2(3) = -5
3=1+2
-5 = -5
3=3
Substitution
● Solve the system: 2p + 3q = 2
p - 3q = -17
● Notice that neither equation is solved for a variable. Since p in the
second equation does not have a coefficient, it will be easier to solve.
p - 3q = -17
p = 3q – 17
● Substitute the value of p into the first equation, and solve.
2p + 3q = 2
2(3q – 17) + 3q = 2
6q – 34 + 3q = 2
9q – 34 = 2
9q = 36
q=4
Substitution
● Substitute the value of q into the second equation to find p.
p = 3q – 17
p = 3(4) – 17
p = -5
● The solution is (-5, 4). (List p first since it comes first alphabetically.)
● Let’s check the solution:
2p + 3q = 2
2(-5) +3(4) = 2
-10 + 12 = 2
2 = 2
p – 3q = -17
-5 - 3(4) = -17
-5 - 12 = -17
-17 = -17 
Substitution
Solve the systems by substitution:
1.
x=4
2x - 3y = -19
2.
3x + y = 7
4x + 2y = 16
3.
2x + y = 5
3x – 3y = 3
4.
2x + 2Y = 4
x – 2y = 0