If the system of equations x+y+z=6,2x+5y+alph | vedclass

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(C) The given system of equations is:$x+y+z=6$  $(1)$$2x+5y+\alpha z=\beta$  $(2)$$x+2y+3z=14$  $(3)$For the system to have infinitely many solutions,

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(C) The given system of equations is:$x+y+z=6$  $(1)$$2x+5y+\alpha z=\beta$  $(2)$$x+2y+3z=14$  $(3)$For the system to have infinitely many solutions,the determinant of the coefficient matrix must be zero,and the augmented matrix must satisfy the consistency condition.Let $D = \begin{vmatrix} 1 & 1 & 1 \ 2 & 5 & \alpha \ 1 & 2 & 3 \end{vmatrix} = 0$.$1(15-2\alpha) - 1(6-\alpha) + 1(4-5) = 0$$15-2\alpha - 6 + \alpha - 1 = 0$$8 - \alpha = 0 \Rightarrow \alpha = 8$.Now,substitute $\alpha = 8$ into the system and use the condition for infinite solutions. From $(1)$ and $(3)$:$x+y = 6-z$$x+2y = 14-3z$Subtracting the first from the second: $y = (14-3z) - (6-z) = 8-2z$.Substituting $y$ into $x+y = 6-z$: $x = 6-z - (8-2z) = z-2$.Substitute $x, y, z$ into $(2)$:$2(z-2) + 5(8-2z) + 8z = \beta$$2z - 4 + 40 - 10z + 8z = \beta$$36 = \beta$.Thus,$\alpha + \beta = 8 + 36 = 44$.

If the system of equations $x+y+z=6$,$2x+5y+\alpha z=\beta$,and $x+2y+3z=14$ has infinitely many solutions,then $\alpha+\beta$ is equal to.

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