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Can you manipulate the coefficients of any variable terms such that they could be eliminated?
(1,-4,3)
The given system consists of equations of planes. Let's use the Elimination Method to find a solution to this system. 3x+3y+6z=9 & (I) 2x+y+3z=7 & (II) x+2y-z=-10 & (III) Currently, none of the terms in this system will cancel out. However, if we multiply (III) by 3 the coefficient of z in this equation will be the additive inverse of the coefficient of z in the second equation; they will add to be 0. 3x+3y+6z=9 2x+y+3z=7 3(x+2y-z)=3(-10) ⇓ 3x+3y+6z=9 2x+y + 3z=7 3x+6y - 3z=-30 We can start by adding the third equation to the second equation to eliminate the z-terms.
(II): Add (III)
(II): Remove parentheses
(II): Add and subtract terms
(III): LHS * 2=RHS* 2
(I): Subtract (III)
(I): Remove parentheses
(I):Add and subtract terms
(I): .LHS /3.=.RHS /3.
Next, we 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.
(I): LHS * (-5)=RHS* (-5)
(II):LHS * 3=RHS* 3
(II): Add (I)
(II): Remove parentheses
(II): Add and subtract terms
(II): .LHS /(-4).=.RHS /(-4).
Now that we know that y=-4, we can substitute it into the first equation to find the value of x.
(I): y= -4
(I): - a(- b)=a* b
(I): LHS-100=RHS-100
(I): .LHS /(-15).=.RHS /(-15).
The value of x is 1. Let's substitute both values into the third equation to find z.
The solution to the system is ( 1, -4, 3). This is the singular point at which all three planes intersect. Now, we can check our solution by substituting the values into the system.
(I), (II), (III): Substitute values
(I), (II), (III): Multiply
(I), (II), (III): Add and subtract terms
Since the substitution of our answers into the given equations resulted in three identities, we know that our solution is correct.