Question

Determine the number of solutions for the following system of linear equations. If there is only onesolution, find the solution.6x - 6y – 2z = 3- 2x + 5y – z = 35x + y - 5z = - 4AnswerKeypadKeyboard ShortcutsSelecting an option will enable input for any required text boxes. If the selected option does not have anyassociated text boxes, then no further input is required.O No SolutionO Only One SolutionX=y =2 =O Infinitely Many Solutions

Answer 1

Explanation

Given the equations;

$\begin{bmatrix}6x-6y-2z=3-----i \\ -2x+5y-z=3----ii \\ 5x+y-5z=-4----iii\end{bmatrix}$

We are asked to find its solution.

First, we isolate x from the first equation

$\begin{gathered} 6x-6y-2z=3 \\ 6x=3+6y+2z \\ x=\frac{3+6y+2z}{6} \end{gathered}$

Substitute x in the remaining equations

$\begin{bmatrix}-2\cdot\frac{3+6y+2z}{6}+5y-z=3 \\ 5\cdot\frac{3+6y+2z}{6}+y-5z=-4\end{bmatrix}=\begin{bmatrix}\frac{9y-5z-3}{3}=3----iv \\ \frac{36y-20z+15}{6}=-----v\end{bmatrix}$

From equation iv above, we will isolate y

$\begin{gathered} \frac{9y-5z-3}{3}=3 \\ 9y-5z-3=9 \\ 9y=9+5z+3 \\ y=\frac{12+5z}{9} \end{gathered}$

Substitute y in equation v

$\begin{gathered} \begin{bmatrix}\frac{36\cdot\frac{5z+12}{9}-20z+15}{6}=-4\end{bmatrix}\Rightarrow\frac{20z+48-20z+15}{6}=-4 \\ \Rightarrow\frac{63}{6}=-4 \\ \Rightarrow\frac{21}{2}=-4 \\ \frac{21}{2}=-4\: \mathrm{\: is\: false,\: therefore\: the\: system\: of\: equations\: has\: no\: solution} \end{gathered}$

Answer: No solution