Question

If 2x+ 3y+5z=50, 3x+ 2y+3z=35, 5x+y+2z= 27 What is the value of 'y'?

Answer 1

Answer:

Step-by-step explanation:

solved using a matrix by using cramer's rule

x=2, y=7, z=5

Answer 2

The value of y is 7.

Given :

2x+ 3y+5z=50,

3x+ 2y+3z=35,

5x+y+2z= 27

To Find:

the value of 'y'.

Solution:

Write the given equations in matrix form

$\left[\begin{array}{ccc}2&3&5\\3&2&3\\5&1&2\end{array}\right]$ $*\left[\begin{array}{ccc}x\\y\\z\end{array}\right]= \left[\begin{array}{ccc}50\\35\\27\end{array}\right]$

According to gaussian elimination method,

$\left[\begin{array}{ccc}2&3&5\\0&-5/2&-9/2\\5&1&2\end{array}\right]*\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\left= \left[\begin{array}{ccc}50\\-40\\27\end{array}\right]$

$\left[\begin{array}{ccc}2&3&5\\0&-5/2&-9/2\\0&-13/2&-21/2\end{array}\right]*\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\left= \left[\begin{array}{ccc}50\\-40\\-98\end{array}\right]$

$\left[\begin{array}{ccc}2&3&5\\0&-5&-9\\0&-13&-21\end{array}\right]*\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\left= \left[\begin{array}{ccc}50\\-80\\-196\end{array}\right]$

$\left[\begin{array}{ccc}2&3&5\\0&-5&-9\\0&0&12/5\end{array}\right]*\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\left= \left[\begin{array}{ccc}50\\-80\\12\end{array}\right]$

$\left[\begin{array}{ccc}2&3&5\\0&-5&-9\\0&0&12\end{array}\right]*\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\left= \left[\begin{array}{ccc}50\\-80\\60\end{array}\right]$

From the above matrix, we can say, 12z = 60 -----(1)

z = 5

5y + 9z = 80  --------(2)

substituting the z value in equation (2)

5y = 80 - 45

    = 35

y = 7

Therefore, The value of y is 7.

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