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3 by 3 Systems of equations
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Solving systems of equations in three unknowns by substitution, elimination or calculator methods, and determining whether the system has no solution, a unique solution or infinitely many solutions.
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Systems of equations with 2 unknowns
SL Core 2.1
We can solve a system of 2 equations and 2 unknowns with different methods.
{3y+2x−2=03y−3x+1=0
By substitution
Rearranging
3y+2x−2=0⇒y=−32x+32
Substituting this into 3y−3x+1=0:
(−2x+2)−3x+1=0
−5x=−3
So x=53, which implies y=−32⋅53+32=154. So the intersection is (53,154).
By elimination
We can eliminate y from the equations by subtracting the second from the first:
(3y+2x−2)−(2y−3x+1)2x−2+3x−15x=0=0=3
So x=53⇒y=154 and the intersection is again (53,154).
We can use either of these methods to systems of equations with 2 equations and 2 unknowns.
Solving systems of equations with 3 unknowns
AHL AA 1.16
Systems of 3 equations with 3 unknowns, for example
(1)(2)(3)⎩⎪⎨⎪⎧2x−3y+4z=85x+2y−z=63x+4y+2z=17
can be solved with a calculator, or by using substitution.
Worked solution
For the system above, equation (2) is a convenient place to start, because it contains just −z, so it is easy to rearrange for z:
z=5x+2y−6
Now substitute this expression for z into equations (1) and (3).
2x−3y+4(5x+2y−6)=8
22x+5y=32
Using equation (3):
3x+4y+2(5x+2y−6)=17
13x+8y=29
We know have a system of 2 equations with 2 unknowns. Eliminate y by taking
8(22x+5y)−5(13x+8y)=8⋅32−5⋅29
176x+40y−65x−40y=256−145
111x=111
x=1
Now substitute x=1 into 13x+8y=29:
13(1)+8y=29
y=2
Finally, substitute x=1 and y=2 into z=5x+2y−6:
z=5(1)+2(2)−6
z=3
So the solution is x=1,y=2,z=3.
Solution count for 3 by 3 systems of equations
AHL AA 1.16
A system of 3 equations with 3 unknowns can have
no solutions
a unique solution
infinitely many solutions
Example
Consider the system of equations
(1)(2)(3)⎩⎪⎨⎪⎧x+2y−z=1−x−y−z=−1x+ay+2z=2
Add equations (1) and (2):
y−2z=0
y=2z
Now add equations (2) and (3):
(a−1)y+z=1
Substitute y=2z:
(a−1)(2z)+z=1
(2a−1)z=1
If 2a−1=0, then we can solve for
z=2a−11
and then find y and z. This represents a unique solution.
But if 2a−1=0, ie a=21, then we have 0z=1, to which there are no solutions.
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