The given system consists of equations of planes. Notice that the coefficient of y in the first equation is the additive inverse of the coefficient of y in the second equation; they will add to be 0. Let's use the Elimination Method to find a solution to this system.
3x - y+z=3 & (I) x + y+2z=4 & (II) x+2y+z=4 & (III)
We can start by adding the first equation to the second equation to eliminate the y-terms.
Having eliminated the y-variable from the second equation, we can continue by creating additive inverse coefficients for y in the first and third equations. Then, we can add or subtract these equations to eliminate y from the first equation.
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 it will be similar to when using it in a system with only two variables.
The solution to the system is ( 1, 1, 1). This is the singular point at which all three planes intersect.
Now, we can check our solution by substituting the values into the system.
3x-y+z=3 & (I) x+y+2z=4 & (II) x+2y+z=4 & (III)
(I), (II), (III): Substitute values
3( 1)- 1+ 1? =3 1+ 1+2( 1)? =4 1+2( 1)+ 1? =4
(I), (II), (III): Identity Property of Multiplication
3-1+1? =3 1+1+2? =4 1+2+1? =4
(I), (II), (III): Add and subtract terms
3=3 âś“ 4=4 âś“ 4=4 âś“
Since the substitution of our answers into the given equations resulted in three identities, we know that our solution is correct.
