Question

How do you solve the system `y=1/3x+2, y=-x-2` by graphing?

Answer 1

the answers are `x=-3,y=1`

Explanation:

let,the given equations be `y=1/3x+2->(1)` and
`y=-x-2->(2)` equating the `R.H.S.` part of both the equations we get `1/3x+2=-x-2rArrx+6=-3x-6rArrx+3x=-6-6rArr4x=-12rArrx=-3`
substituting the value of `x` in `eq^n (1)` we get `y=1/3(-3)+2=-1+2=1`