The given system consists of equations of planes. Notice that the coefficient of x in the first equation is the same number as the coefficient of x in the third equation; they will subtract to be 0. Let's use the Elimination Method to find a solution to this system.
x-y+2z=-7 & (I) y+z=1 & (II) x=2y+3z & (III)
We can start by subtracting the third equation from the first equation to eliminate the x-terms.
Next, we will 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 it will be similar to when using it in a system with only two variables.
The solution to the system is ( 0, 3, -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.
x-y+2z=-7 & (I) y+z=1 & (II) x=2y+3z & (III)
(I), (II), (III): Substitute values
0- 3+2( -2)? =-7 3+( -2)? =1 0? =2( 3)+3( -2)
(I), (II), (III): Multiply
0-3-4? =-7 3+(-2)? =1 0? =6-6
(I), (II), (III): Add and subtract terms
-7=-7 âś“ 1=1 âś“ 0=0 âś“
Since the substitution of our answers into the given equations resulted in three identities, we know that our solution is correct.
