How do you solve the system #2x+y=1# and #x+2y=-2# by graphing?
1 Answer
See a solution process below:
Explanation:
For each of the equations solve the equation for two points:
Equation 1:
For
For
Equation 2:
For
For
Graph Equation 1 by plotting the two points and drawing a straight line through them:
graph{(x^2+(y-1)^2-0.03)((x-2)^2+(y+3)^2-0.03)(2x + y - 1)=0}
Add the graph Equation 2 by plotting the two points and drawing a straight line through them:
graph{(2x + y - 1)((x+2)^2+y^2-0.03)(x^2+(y+1)^2-0.03)(x+2y+2)=0}
Solution:
graph{(2x + y - 1)(x+2y+2)((x-1.3333333333)^2+(y+1.666666667)^2-0.0025)=0 [-2, 2, -2, 0]}
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