Find the value of x, y, z for the given system of equations : 2x+3y+2z=50, x+4y+3z=40, 3x+3y+5z=60

The correct option is: x=125/8, y=15/2, z=-(15/8).

To find the values of $x$, $y$, and $z$, we can solve the given system of linear equations step-by-step:

Equation 1:
$2x+3y+2z=50$

Equation 2:
$x+4y+3z=40$

Equation 3:
$3x+3y+5z=60$

Step 1: Eliminate one variable from the equations.
Let us eliminate $x$ using Equation 2. From Equation 2, we can express $x$ in terms of $y$ and $z$:
$x=40-4y-3z$

Step 2: Substitute this expression for $x$ into Equation 1 and Equation 3.

Substituting into Equation 1:
$2(40-4y-3z)+3y+2z=50$
Simplifying:
$80-8y-6z+3y+2z=50$
$80-5y-4z=50$
Subtract 80 from both sides:
$-5y-4z=-30$
Multiplying by -1 gives Equation 4:
$5y+4z=30$

Substituting into Equation 3:
$3(40-4y-3z)+3y+5z=60$
Simplifying:
$120-12y-9z+3y+5z=60$
$120-9y-4z=60$
Subtract 120 from both sides:
$-9y-4z=-60$
Multiplying by -1 gives Equation 5:
$9y+4z=60$

Step 3: Solve the system of two equations (Equation 4 and Equation 5).

Subtract Equation 4 from Equation 5:
$(9y+4z)-(5y+4z)=60-30$
$4y=30$
$y=\frac{30}{4}=\frac{15}{2}$

Step 4: Find $z$ using the value of $y$.
Substitute $y=\frac{15}{2}$ into Equation 4:
$5\left(\frac{15}{2}\right)+4z=30$
$\frac{75}{2}+4z=30$
$4z=30-\frac{75}{2}$
$4z=\frac{60}{2}-\frac{75}{2}$
$4z=-\frac{15}{2}$
$z=-\frac{15}{8}$

Step 5: Find $x$ using the values of $y$ and $z$.
Substitute $y=\frac{15}{2}$ and $z=-\frac{15}{8}$ into Equation 2:
$x+4\left(\frac{15}{2}\right)+3(-\frac{15}{8})=40$
$x+30-\frac{45}{8}=40$
$x=40-30+\frac{45}{8}$
$x=10+\frac{45}{8}$
$x=\frac{80}{8}+\frac{45}{8}$
$x=\frac{125}{8}$

Thus, the final values are $x=\frac{125}{8}$, $y=\frac{15}{2}$, and $z=-\frac{15}{8}$.