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Can you manipulate the coefficients of any variable terms such that they could be eliminated?
(0,3,4)
The given system consists of equations of planes. Notice that the coefficient of z in the first equation is the additive inverse of the coefficient of z in the second equation; they will add to be 0. Let's use the Elimination Method to find a solution to this system. 2x+3y - 2z=1 & (I) - x-y + 2z=5 & (II) 3x+2y-3z=-6 & (III) We can start by adding the second equation to the first equation to eliminate the z-terms.
(I): Add (II)
(I): Remove parentheses
Add and subtract terms
Having eliminated the z-variable from the first equation, we can continue by creating additive inverse coefficients for z in the second and third equations. Then, we can add or subtract these equations to eliminate z from the second equation.
(II): LHS * 3=RHS* 3
(III): LHS * 2=RHS* 2
(II):Add (III)
(II): Remove parentheses
(II):Add and subtract terms
Next, we will use our two equations that are only in terms of x and y to solve for the value of one of the variables. We will once again apply the Elimination Method, but this time it will be similar to when using it in a system with only two variables.
Now that we know that x=0, we can substitute it into the second equation to find the value of y.
(II): x= 0
(II): Zero Property of Multiplication
(II): .LHS /(-2).=.RHS /(-2).
The value of y is 3. Let's substitute both values into the third equation to find z.
The solution to the system is (0,3,4). This is the singular point at which all three planes intersect.