Answer to Question #104267 in Algebra for Gayatri Yadav

Question

The system of equations 2x + 3y + 5z = 1 and 3x + 2y + z = 5 has a unique solution.

Answer 1

The given system of equations is :

2x + 3y + 5z = 1 -> eq. (1)

3x + 2y + z = 5 -> eq. (2)

Multiplying eq. (1) with 3 and eq. (2) with 2, we get :

6x + 9y + 15z = 3 -> eq. (3)

and, 6x + 4y + 2z = 10 -> eq. (4)

respectively.

Subtracting eq. (4) from eq. (3) :

5y + 13z = -7 -> eq. (5)

Multiplying eq. (1) with 2 and eq. (2) with 3, we get :

4x + 6y + 10z = 2 -> eq. (6)

and, 9x + 6y + 3z = 15 -> eq. (7)

respectively.

Subtracting eq. (7) from eq. (6) :

-5x + 7z = -13 -> eq. (8)

Let z = t, where t is a parameter.

We can write eq. (5) and eq. (8) as :

5y + 13t = -7 & -5x + 7t = -13

i.e., 5y = -7 - 13t & 5x = 13 + 7t

y = -(7/5) - (13t/5) & x = (13/5) + (7t/5)

Therefore, the solution for the given system of equations is :

x = (13/5) + (7t/5)

y = -(7/5) - (13t/5)

z = t

where, t can take any value.

Hence, the given system of equations has infinite number of solutions.

Therefore, the given statement is incorrect.