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Can you manipulate the coefficients of any variable terms such that they could be eliminated?
(2,1,-5)
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-y + z=-2 & (I) x+3y - z=10 & (II) x+2z=-8 & (III) We can start by adding the second equation to the first equation to eliminate the z-terms.
(I): Add (II)
(I): Remove parentheses
(I): Add 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 * 2=RHS* 2
(II): Add (III)
(II): Remove parentheses
(II):Add 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 y=1, we can substitute it into the first equation to find the value of x.
(I): y= 1
(I): Identity Property of Multiplication
(I): LHS-2=RHS-2
(I): .LHS /3.=.RHS /3.
The value of x is 2. Let's substitute the value of x into the third equation to find z.
The solution to the system is (2,1,-5). This is the singular point at which all three planes intersect.