Question

The following system of linear equations :
2x + 3y + 2z = 9 , 3x + 2y + 2z = 9 and x – y + 4z = 8
(a) Does not have any solution
(b) Has a unique solution
(c) Has a solution $\alpha +{{\beta }^{2}}+{{\gamma }^{3}}$$\left( \alpha ,\beta ,\gamma \right)$satisfying $\alpha +{{\beta }^{2}}+{{\gamma }^{3}}$ = 12
 (d) Has infinitely many solutions

Answer 1

Hint: In this, we have given three equations. Now we solve the equations to get the value of x, y, and z. After getting the value of x, y and z we get to know whether it is unique, infinitely, and does not have a solution.

Complete step by step Solution:
We have given the equation
2x + 3y + 2z = 9 . . . . . . . . . . . . . . . . . . . . . . . .. . . .. (1)
3x + 2y + 2z = 9 . . . . . .. . .. .. .. . . . . . . .. …. . . .. . . .. ..(2)
 x – y + 4z = 8 . . . . . . …. . . . . . . . . . . . . . . . . . . . . . . . . .(3)
now subtracting equation (2) from equation (1), we get
-( x – y ) = 0
i.e. x – y = 0 . . . . . . . . .. . . . . . . . . .. . . . . . . . . . . . . . . . .(4)
Now we put x – y = 0 in equation (3), we get
0 + 4z = 8
i.e. z = 2
Now put z = 2 in equation (1), we get
2x + 3y + 2(2) = 9
I.e. 2x + 3y + 4 = 9
i.e. 2x + 3y = 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(5)
now we solve equation (4) and ( 5) by substitution method
from equation (4) , x – y = 0
i.e. x = y
put x = y in equation (5), we get
2x + 3x = 5
i.e. 5x = 5
i.e. x = 1
if x = 1, then y is also equal to 1
Therefore, the system has a unique solution.

Hence, the correct option is b.

Note: Students made mistakes to get the value of x, y, and z. There are many methods to solve the equations like elimination and substitution methods. You can solve the question using the method which you know. There is no hard and fast rule in solving equations.