How To Solve A System Of Equations With 3 Variables?
To solve a system of three equations with three variables you can use substitution, linear combinations, or any number of matrix applications including Cramer's method and Gaussian elimination.
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Suppose we are asked to solve a linear system of three equations in three variables. If a unique solution exists it will be an ordered three-tuple (e.g. (x,y,z) ) which represents a point in Euclidean 3-space. The system will be said to be consistent and independent.
If the solution is not unique we can present the solution as an ordered three-tuple with each element written in terms of one variable. E.g. the solution might appear as (x, x+2, -2x). The system is said to be consistent and dependent.
It is also possible that the system has no solution in which case it is said to be inconsistent.
Graphically a system that is consistent and independent is the point of intersection of three planes (like the corner of a room.) If the system is consistent and dependent the solution is the intersection of three planes in a line (much like the spine of a book.)
To solve we can use techniques we already know from systems of two equations in two unknowns.
Example: Solve the following system:
`x+2y+3z=2`
`3x-5y-4z=15`
`-2x-3y+2z=2`
A. We can use substitution. Choose an equation and a variable to solve for. In this case it would be simple to choose to solve the first equation for x.
`x=-2y-3z+2` Now we substitute this expression for x in the remaining equations.
`3(-2y-3z+2)-5y-4z=15`
`-2(-2y-3z+2)-3y+2z=2` Simplifying we get:
`-11y-13z=9`
`y+8z=6`
Now we have two equations in two unknowns and we have many techniques to solve this system. Solving the second equation for y and substituting we get
`-11(-8z+6)-13z=9 ==> 75z=75 ==> z=1`
Now we can back substitute to get y then x:
`y+8=6 ==> y=-2`
`x=-2(-2)-3(1)+2=3`
The solution is (3,-2,1) which we would check in each of the three original equations.
B. We can also use linear combinations:
`x+2y+3z=2`
`3x-5y-4z=15`
`-2x-3y+2z=2`
Choose to eliminate x; if the equations are marked i, ii, iii then take 3i-ii to get 11y+13z=-9
Then 2i+iii to get y+8z=6
Now we have the system:
11y+13z=-9
y+8z=6
Eliminating y we get 75z=75 as above.
There are many techniques using matrices (Cramer's method, Gaussian elimination, matrix multiplication etc...) as well as some linear algebra techniques with basis vectors.
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