Question

System of Equations Need help Trig

Answer 1

Looks like the system is

$\begin{cases}-2x+y+6z=1\\3x+2y+5z=16\\7x+3y-4z=11\end{cases}$

We can eliminate $y$ by taking

$(3x+2y+5z)-2(-2x+y+6z)=16-2(1)$

$\implies3x+2y+5z+4x-2y-12z=16-2$

$\implies7x-7z=14$

$\implies x-z=2$

so that $z=x-2$, and

$(7x+3y-4z)-3(-2x+y+6z)=11-3(1)$

$\implies7x+3y-4z+6x-3y-18z=11-3$

$\implies13x-22z=8$

Substitute $z=x-2$ into this last equation and solve for $x$:

$13x-22(x-2)=8$

$\implies13x-22x+44=8$

$\implies-9x=-36$

$\implies x=4$

Then

$z=x-2$

$\implies z=4-2$

$\implies z=2$

Plug these values into any one of the original equation to solve for $y$:

$-2x+y+6z=1$

$\implies-2(4)+y+6(2)=1$

$\implies-8+y+12=1$

$\implies y=-3$

Hence the solution is x = 4, y = -3, and z = 2.