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Can you manipulate the coefficients of any variable terms such that they could be eliminated?
(3,2,-3)
The given system consists of equations of planes. Notice that the coefficient of y in the second equation is the additive inverse of the coefficient of y in the third equation; they will add to be 0. Therefore, let's use the Elimination Method to find a solution to this system. x+2y+z=4 & (I) 2x - y+4z=-8 & (II) -3x + y-2z=-1 & (III) We can start by adding the second equation to the third equation to eliminate the y-terms.
(III): Add (II)
(III): Remove parentheses
(III): Add and subtract terms
Having eliminated the y-variable from the third equation, we can continue by creating additive inverse coefficients for y in the first and second equations. Then, we can add or subtract these equations to eliminate y from the second equation.
(II): LHS * 2=RHS* 2
(II): Add (I)
(II): Remove parentheses
(II): Add terms
Next, we use our two equations that are only in terms of x and z to solve for the value of one of the variables. We will once again apply the Elimination Method, but this time will be similar to when using it in a system with only two variables.
Now that we know that z=-3, we can substitute it into the third equation to find the value of x.
(III): z= -3
(III): a(- b)=- a * b
(III): LHS+30=RHS+30
(III): .LHS /(-5).=.RHS /(-5).
The value of x is 3. Let's substitute both values into the first equation to find y.
The solution to the system is (3,2,-3). This is the singular point at which all three planes intersect.