Sign In
Can you manipulate the coefficients of any variable terms such that they could be eliminated?
(2,0,- 2)
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 third equation; they will add to be 0. Let's use the Elimination Method to find a solution to this system. 7x-2y-5z=24 & (I) - x+3y+4z=- 10 & (II) x-y-z=4 & (III) ⇓ 7x-2y-5z=24 & (I) - 1x+3y+4z=- 10 & (II) 1x-y-z=4 & (III) We can start by adding the third equation to the second equation to eliminate the x-terms.
(II): Add (III)
(II): Add and subtract terms
(III): LHS * (- 7)=RHS* (- 7)
(III): Add (I)
(III): Add and subtract terms
Next, we use our two equations that are only in terms of y 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.
(II): LHS * (- 2)=RHS* (- 2)
(III): LHS * 3=RHS* 3
(II): Add (III)
(II): a+(- b)=a-b
(II): Add and subtract terms
(II): .LHS /11.=.RHS /11.
Now that we know that y=0, we can substitute it into the third equation to find the value of z.
(III): y= 0
(III): Zero Property of Multiplication
(III): Identity Property of Addition
(III): .LHS /6.=.RHS /6.
The value of z is - 2. Let's substitute both values into the first equation to find x.
(III): y= 0, z= - 2
(I): Zero Property of Multiplication
(I): - a(- b)=a* b
(I): Add and subtract terms
(I): LHS-10=RHS-10
(I): .LHS /7.=.RHS /7.
The solution to the system is ( 2, 0, - 2). 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.