Question

Solve x+y+z=7,x+2y+3z=16,x+3y+4z=22 answer

Answer 1

Given:

Solve x+y+z=7,x+2y+3z=16,x+3y+4z=22.

To Solve:

solve for x?

Solution:

it is given that-

x+y+z=7,x+2y+3z=16,x+3y+4z=22

then , the augmented matrix is

$\[\left[ {\begin{array}{*{20}{c}} 1&1&1\vline & 7 \\ 1&2&3\vline & {16} \\ 1&3&4\vline & {22} \end{array} } \right]\]$

On solving , we get

$\left[ {\begin{array}{*{20}{c}} 1&1&1\vline & 7 \\ 0&1&2\vline & 9 \\ 0&2&3\vline & {15} \end{array} } \right]{R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1} \hfill$

$\left[ {\begin{array}{*{20}{c}} 1&1&1\vline & 7 \\ 0&1&2\vline & 9 \\ 0&0&1\vline & 3 \end{array} } \right]{R_3} \to 2{R_2} - {R_3} \hfill$

$\left[ {\begin{array}{*{20}{c}} 1&1&0\vline & 4 \\ 0&1&0\vline & 3 \\ 0&0&1\vline & 3 \end{array} } \right]{R_1} \to {R_1} - {R_3},{R_2} \to {R_2} - 2{R_3} \hfill$

$\left[ {\begin{array}{*{20}{c}} 1&0&0\vline & 1 \\ 0&1&0\vline & 3 \\ 0&0&1\vline & 3 \end{array} } \right]{R_1} \to {R_1} - {R_2} \hfill$

Hence, on solving above matrix , we get

$x=1,y=3,z=3$.

Answer 2

Answer:

Solve x+y+z=7,x+2y+3z=16,x+3y+4z=22 answer