Question

I was never good with fractions, could someone help?

Answer 1

Answer:

Thus, the answer is:

$\left \{ {{x=4} \atop {y=2}} \ \atop {z=-4}} \right$

Step-by-step explanation:

Considering the system of equations

$\left \{ {{x+y+z=2} \atop {4x+y+z=14}} \ \atop {x-y+4z=-14}} \right\}$

Rewrite the system in matrix form and solve it by Gaussian Elimination (Gauss-Jordan elimination)

$\left \{ {{x+y+z=2} \atop {4x+y+z=14}} \ \atop {x-y+4z=-14}} \right\}\rightarrow \left[\begin{array}{ccc}1&1&1:2\\4&1&1:14\\1&-1&4:-14\end{array}\right]$

R2 - 4 R1 → R2 (multiply 1 row by 4 and subtract it from 2 row); R3 - 1 R1 → R3 (multiply 1 row by 1 and subtract it from 3 row)

  $\left[\begin{array}{ccc}1&1&1:2\\0&-3&-3:6\\0&-2&3:-16\end{array}\right]$

R2 / -3 → R2 (divide the 2 row by -3)

$\left[\begin{array}{ccc}1&1&1:2\\0&1&1:-2\\0&-2&3:-16\end{array}\right]$

R1 - 1 R2 → R1 (multiply 2 row by 1 and subtract it from 1 row); R3 + 2 R2 → R3 (multiply 2 row by 2 and add it to 3 row)

$\left[\begin{array}{ccc}1&0&0:4\\0&1&1:-2\\0&0&5:-20\end{array}\right]$

R3 / 5 → R3 (divide the 3 row by 5)

$\left[\begin{array}{ccc}1&0&0:4\\0&1&1:-2\\0&0&1:-4\end{array}\right]$

R2 - 1 R3 → R2 (multiply 3 row by 1 and subtract it from 2 row)

$\left[\begin{array}{ccc}1&0&0:4\\0&1&0:2\\0&0&1:-4\end{array}\right]$

Thus, the answer is:

$\left \{ {{x=4} \atop {y=2}} \ \atop {z=-4}} \right$

Make a check:

4 + 2 + (-4) = 4 + 2 - 4 = 2

4·4 + 2 + (-4) = 16 + 2 - 4 = 14

4 - 2 + 4·(-4) = 4 - 2 - 16 = -14

Check completed successfully.