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.
We can solve a system of 2 equations and 2 unknowns with different methods.
Rearranging
Substituting this into 3y−3x+1=0:
So x=53, which implies y=−32⋅53+32=154. So the intersection is (53,154).
We can eliminate y from the equations by subtracting the second from the first:
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.
Systems of 3 equations with 3 unknowns, for example
can be solved with a calculator, or by using substitution.
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:
Now substitute this expression for z into equations (1) and (3).
Using equation (3):
We know have a system of 2 equations with 2 unknowns. Eliminate y by taking
Now substitute x=1 into 13x+8y=29:
Finally, substitute x=1 and y=2 into z=5x+2y−6:
So the solution is x=1,y=2,z=3.
A system of 3 equations with 3 unknowns can have
no solutions
a unique solution
infinitely many solutions
Consider the system of equations
Add equations (1) and (2):
Now add equations (2) and (3):
Substitute y=2z:
If 2a−1=0, then we can solve for
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.
Nice work completing 3 by 3 Systems of equations, here's a quick recap of what we covered:
Exercises checked off
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.
We can solve a system of 2 equations and 2 unknowns with different methods.
Rearranging
Substituting this into 3y−3x+1=0:
So x=53, which implies y=−32⋅53+32=154. So the intersection is (53,154).
We can eliminate y from the equations by subtracting the second from the first:
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.
Systems of 3 equations with 3 unknowns, for example
can be solved with a calculator, or by using substitution.
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:
Now substitute this expression for z into equations (1) and (3).
Using equation (3):
We know have a system of 2 equations with 2 unknowns. Eliminate y by taking
Now substitute x=1 into 13x+8y=29:
Finally, substitute x=1 and y=2 into z=5x+2y−6:
So the solution is x=1,y=2,z=3.
A system of 3 equations with 3 unknowns can have
no solutions
a unique solution
infinitely many solutions
Consider the system of equations
Add equations (1) and (2):
Now add equations (2) and (3):
Substitute y=2z:
If 2a−1=0, then we can solve for
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.
Nice work completing 3 by 3 Systems of equations, here's a quick recap of what we covered:
Exercises checked off