This is the system I need to solve :
$$ \begin{cases} 10-14x+8y+5z=0 \\ 10-x-2y-6z=0 \\ 10-9x+12y-2z=0 \\ \end{cases} $$
How do I solve for all three?
Any help would be of great help. But I need step by step instructions for each unknown. Thanks
This is the system I need to solve :
$$ \begin{cases} 10-14x+8y+5z=0 \\ 10-x-2y-6z=0 \\ 10-9x+12y-2z=0 \\ \end{cases} $$
How do I solve for all three?
Any help would be of great help. But I need step by step instructions for each unknown. Thanks
On
Well you should try eliminating x from the first and second equations by multiplying them with a suitable integer and then adding or subtracting the two equations as required to eliminate x. This would give you an equation in y and z. Then do the same with the second and third equations to eliminate x. Now you two simultaneous equations in y and z . Solve them by substitution of any other method for your choice to obtain the values of y and z. Then put this value in any one of the three equations to find the value of X. You will get the answer!
wrting your system in the form $$-x-2y-6z=-10$$ $$-9x+12y-2z=-10$$ $$-14x+8y+5z=-10$$ then we get multiplying the first equation by $-9$ and adding to the second : $$30y+52z=80$$ and multiplying the first by $-14$ and adding to the third: $$36y+89z=130$$ can you do the last steps?