Question

What is the solution to the system?

-2x + y + 6z = 1
3x + 2y + 5z = 16
7x + 3y – 4z = 11

Answer 1

x = 4, y = -2, z = 3

Explanation:

`\left\{\begin{array}{ccc}-2x+y+6z=1&(1)\\3x+2y+5z=16&(2)\\7x+3y-4z=11&(3)\end{array}\right\\\\(1)\\-2x+y+6z=1\qquad\text{add}\ 2x\ \text{to both sides}\\y+6z=2x+1\qquad\text{subtract}\ 6z\ \text{from both sides}\\y=2x-6z+1\\\\\text{Substitute it to (2) and (3):}\\\\\left\{\begin{array}{ccc}3x+2(2x-6z+1)+5z=16\\7x+3(2x-6z+1)-4z=11\end{array}\right\qquad\text{use the distributive property}\\\\`

`\left\{\begin{array}{ccc}3x+4x-12z+2+5z=16&\text{subtract 2 from both sides}\\7x+6x-18z+3-4z=11&\text{subtract 3 from both sides}\end{array}\right\\\left\{\begin{array}{ccc}7x-7z=14&\text{divide both sides by 7}\\13x-22z=8\end{array}\right\\\left\{\begin{array}{ccc}x-z=2&\text{multiply both sides by (-13)}\\13x-22z=8\end{array}\right`

`\underline{+\left\{\begin{array}{ccc}-13x+13z=-26\\13x-22z=8\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad-9z=-18\qquad\text{divide both sides by (-2)}\\.\qquad\qquad \boxed{z=2}\\\\\text{Put it ot the equation}\ x-z=2:\\x-2=2\qquad\text{add 2 to both sides}\\\boxed{x=4}\\\\\text{Put the values of}\ x\ \text{and}\ z\ \text{to the equation}\ y=2x-6z+1:\\\\y=2(4)-6(2)+1\\y=8-12+1\\\boxed{y=-3}`