How do you solve this system of equations: \[x + 2y + z = 4\] , \[x + y + z = 3\] ; \[x - 9y + 3z = - 5\] ?

Hint: Here in this question, given the system of linear equations. We have to find the unknown values that are \[x\] , \[y\] and \[z\] solving these equations by using the elimination method. In elimination methods either we add or subtract the equations to find the unknown values of \[x\] and \[y\] .

Complete step by step solution:
Let us consider the equation and we will name it as (1), (2) and (3)
 \[x + 2y + z = 4\] ----------(1)
 \[x + y + z = 3\] -----------(2)
 \[x - 9y + 3z = - 5\] --------(3)
Now we have to solve these three equations to find the unknown. Firstly we have to compare all three equations.
Subtract equation (2) by equation (1), we have
 \[ \Rightarrow \,\,x + 2y + z - \left( {x + y + z} \right) = 4 - 3\]
 \[ \Rightarrow \,\,x + 2y + z - x - y - z = 4 - 3\]
On simplification, we get
 \[ \Rightarrow \,\,y = 1\]
Substitute value \[y = 1\] in all three equation, then we have
Equation (1) can be written as:
 \[ \Rightarrow \,\,x + 2\left( 1 \right) + z = 4\]
 \[ \Rightarrow \,\,x + 2 + z = 4\]
 \[ \Rightarrow \,\,x + z = 4 - 2\]
 \[ \Rightarrow \,\,x + z = 2\] ----------(4)
Equation (2) can be written as?
 \[ \Rightarrow \,x + 1 + z = 3\]
 \[ \Rightarrow \,x + z = 3 - 1\]
 \[ \Rightarrow \,x + z = 2\] --------(5)
Equation (3) can be written as
 \[ \Rightarrow \,x - 9\left( 1 \right) + 3z = - 5\]
 \[ \Rightarrow \,x - 9 + 3z = - 5\]
 \[ \Rightarrow \,x + 3z = - 5 + 9\]
 \[ \Rightarrow \,x + 3z = 4\] ---------(6)
Subtract equation (6) from equation (5), then we have
 \[ \Rightarrow \,x + 3z - \left( {x + z} \right) = 4 - 2\]
 \[ \Rightarrow \,x + 3z - x + - z = 4 - 2\]
On simplification, we get
 \[ \Rightarrow \,2z = 2\]
Divide both side by 2, then
 \[ \Rightarrow \,z = 1\]
We have found the value of \[y\] and \[z\] now we have to find the value of \[x\] . so we will substitute the value of \[y\] and \[z\] now to any one of the equation (1) or (2) or (3) . we will substitute the value of \[y\] and \[z\] in equation (2).
Consider equation (2)
 \[ \Rightarrow \,\,x + y + z = 3\]
Substitute values of \[y\] and \[z\] , then we have
 \[ \Rightarrow \,\,x + 1 + 1 = 3\]
 \[ \Rightarrow \,\,x + 2 = 3\]
 \[ \Rightarrow \,\,x = 3 - 2\]
On simplification, we get
 \[ \Rightarrow \,\,x = 1\]
Hence we got the unknown values \[x = 1\] , \[y = 1\] and.
We can check whether these values are correct or not by substituting the unknown values in the given equations and we have to prove L.H.S is equal to R.H.S
Now we will substitute the value of \[x\] , \[y\] and \[z\] in equation (1) so we have
 \[x + 2y + z = 4\]
 \[ \Rightarrow \,1 + 2\left( 1 \right) + 1 = 4\]
 \[ \Rightarrow \,1 + 2 + 1 = 4\]
 \[ \Rightarrow \,\,\,4 = 4\]
 \[\therefore \,\,\,LHS = RHS\]
Hence the values of the unknown that are \[x\] \[y\] and \[z\] , are the correct values which satisfy the equation.
So, the correct answer is “x = 1, y = 1 and z = 1”.

Note: In this type of question while eliminating the term we must be aware of the sign where we change the sign by the alternate sign. In this we have a chance to verify our answers. In the elimination method we have made the one term have the same coefficient such that it will be easy to solve the equation.