How do you solve #y=x+4#, #y=-2x+1# by graphing?
1 Answer
See a solution process below:
Explanation:
First graph the equation
Graph Equation 1:
First graph the equation
For
For
graph{(x^2+(y-4)^2-0.125)((x-4)^2+(y-8)^2-0.125)(y-x-4)=0 [-20, 20, -10, 10]}
Graph Equation 2:
We do the same for the second equation:
For
For
graph{((x+3)^2+(y-7)^2-0.125)(y+2x-1)(x^2+(y-1)^2- 0.125)(x^2+(y-4)^2-0.125)((x-4)^2+(y-8)^2-0.125)(y-x-4)=0 [-20, 20, -10, 10]}
The two lines intersect at:
graph{((x+1)^2+(y-3)^2-0.075)(y+2x-1)(y-x-4)=0 [-10, 10, -5, 5]}