Mason's question via Facebook about solving a system of equations

[Prove It]

Solve the following system for $\displaystyle \begin{align*} x, y, z \end{align*}$:

$\displaystyle \begin{align*} 5\,x - 2\,y + z &= 3 \\ 3\,x + y + 3\,z &= 5 \\ 6\,x + y - 4\,z &= 62 \end{align*}$

The LCM of the $\displaystyle \begin{align*} x \end{align*}$ coefficients is 30, so multiplying the first equation by 6, the second by 10 and the third by 5 gives

$\displaystyle \begin{align*} 30\,x - 12\,y + 6\,z &= 18 \\ 30\,x + 10\,y + 30\,z &= 50 \\ 30\,x + 5\,y - 20\,z &= 310 \end{align*}$

Applying R2 - R1 to R2 and R3 - R1 to R3 we have

$\displaystyle \begin{align*} 30\,x + 12\,y - 6\,z &= 18 \\ 22\,y + 24\,z &= 32 \\ 17\,y - 26\,z &= 292 \end{align*}$

Dividing the second equation by 2 gives

$\displaystyle \begin{align*} 30\,x + 12\,y - 6\,z &= 18 \\ 11\,y + 12\,z &= 16 \\ 17\,y - 26 \,z &= 292 \end{align*}$

The LCM of the y coefficients in rows 2 and 3 is 187, so multiplying the second equation by 17 and the third equation by 11 we have

$\displaystyle \begin{align*} 30\,x + 12\,y - 6\,z &= 18 \\ 187\,y + 204\,z &= 272 \\ 187\,y - 286\,z &= 3\,212 \end{align*}$

Applying R3 - R2 to R2 we have

$\displaystyle \begin{align*} 30\,x + 12\,y - 6\,z &= 18 \\ 187\,y + 204\,z &= 272 \\ - 490\,z &= 2\,940 \end{align*}$

Since $\displaystyle \begin{align*} -490\,z = 2\,940 \implies z = -6 \end{align*}$, then

$\displaystyle \begin{align*} 187\,y + 204 \, \left( -6 \right) &= 272 \\ 187\,y - 1\,224 &= 272 \\ 187\,y &= 1\,496 \\ y &= 8 \end{align*}$

and

$\displaystyle \begin{align*} 5\,x - 2\,\left( 8 \right) + \left( -6 \right) &= 3 \\ 5\,x - 22 &= 3 \\ 5\,x &= 25 \\ x &= 5 \end{align*}$

So the solution is $\displaystyle \begin{align*} \left( x , y , z \right) = \left( 5, 8, -6 \right) \end{align*}$.

 

[fresh_42]

The LCM of the $\displaystyle \begin{align*} x \end{align*}$ coefficients is 30, so multiplying the first equation by 6, the second by 10 and the third by 5 gives

$\displaystyle \begin{align*} 30\,x - 12\,y + 6\,z &= 18 \\ 30\,x + 10\,y + 30\,z &= 50 \\ 30\,x + 5\,y - 20\,z &= 310 \end{align*}$

Applying R2 - R1 to R2 and R3 - R1 to R3 we have

$\displaystyle \begin{align*} 30\,x + 12\,y - 6\,z &= 18 \\ 22\,y + 24\,z &= 32 \\ 17\,y - 26\,z &= 292 \end{align*}$

Dividing the second equation by 2 gives

$\displaystyle \begin{align*} 30\,x + 12\,y - 6\,z &= 18 \\ 11\,y + 12\,z &= 16 \\ 17\,y - 26 \,z &= 292 \end{align*}$

The LCM of the y coefficients in rows 2 and 3 is 187, so multiplying the second equation by 17 and the third equation by 11 we have

$\displaystyle \begin{align*} 30\,x + 12\,y - 6\,z &= 18 \\ 187\,y + 204\,z &= 272 \\ 187\,y - 286\,z &= 3\,212 \end{align*}$

Applying R3 - R2 to R2 we have

$\displaystyle \begin{align*} 30\,x + 12\,y - 6\,z &= 18 \\ 187\,y + 204\,z &= 272 \\ - 490\,z &= 2\,940 \end{align*}$

Since $\displaystyle \begin{align*} -490\,z = 2\,940 \implies z = -6 \end{align*}$, then

$\displaystyle \begin{align*} 187\,y + 204 \, \left( -6 \right) &= 272 \\ 187\,y - 1\,224 &= 272 \\ 187\,y &= 1\,496 \\ y &= 8 \end{align*}$

and

$\displaystyle \begin{align*} 5\,x - 2\,\left( 8 \right) + \left( -6 \right) &= 3 \\ 5\,x - 22 &= 3 \\ 5\,x &= 25 \\ x &= 5 \end{align*}$

So the solution is $\displaystyle \begin{align*} \left( x , y , z \right) = \left( 5, 8, -6 \right) \end{align*}$.

Correct. The next step would be to write it in form of matrices. 'Gaussian elimination' would be a suitable search key.