\mathrm{Find\; the\; value\; of\; x,y\; and\; z\; form\; the\; following\; equation\; 3}x+2y-z=24,x-5y+3z=16,x+y-3z=21

The correct answer is:
x=\frac{389}{42},y=\frac{-97}{21},z=\frac{-229}{42}

To solve for x, y, and z, we can use the method of elimination to reduce the system of three linear equations into two equations with two variables.

Given system of equations:
(1) 3x+2y-z=24
(2) x-5y+3z=16
(3) x+y-3z=21

Step 1: Eliminate z by combining equations (2) and (3)
Add Equation (2) and Equation (3) directly:
(x-5y+3z)+(x+y-3z)=16+21
2x-4y=37    — (Equation 4)

Step 2: Eliminate z using equations (1) and (3)
Multiply Equation (1) by 3:
3(3x+2y-z)=3\left(24\right)
9x+6y-3z=72    — (Equation 5)
Now subtract Equation (3) from Equation (5):
(9x+6y-3z)-(x+y-3z)=72-21
8x+5y=51    — (Equation 6)

Step 3: Solve for y using Equations (4) and (6)
Multiply Equation (4) by 4 to align the coefficients of x:
4(2x-4y)=4\left(37\right)
8x-16y=148    — (Equation 7)
Subtract Equation (7) from Equation (6):
(8x+5y)-(8x-16y)=51-148
21y=-97
y=\frac{-97}{21}

Step 4: Solve for x
Substitute the value of y into Equation (4):
2x-4\left(\frac{-97}{21}\right)=37
2x+\frac{388}{21}=37
2x=37-\frac{388}{21}=\frac{777-388}{21}=\frac{389}{21}
x=\frac{389}{42}

Step 5: Solve for z
Substitute x and y into Equation (3):
x+y-3z=21
3z=x+y-21
3z=\frac{389}{42}+\left(\frac{-97}{21}\right)-21
Convert to a common denominator of 42:
3z=\frac{389}{42}-\frac{194}{42}-\frac{882}{42}
3z=\frac{389-194-882}{42}=\frac{-687}{42}
Divide by 3:
z=\frac{-687}{126}=\frac{-229}{42}

Thus, the values are:
x=\frac{389}{42},y=\frac{-97}{21},z=\frac{-229}{42}