How do you solve the system #2x+y=1# and #x+2y=-2# by graphing?

1 Answer
Oct 15, 2017

Answer:

See a solution process below:

Explanation:

For each of the equations solve the equation for two points:

Equation 1:

For #x = 0#:

#(2 * 0) + y = 1#

#0 + y = 1#

#y = 1# or #(0, 1)#

For #x = 2#:

#(2 * 2) + y = 1#

#4 + y = 1#

#4 - color(red)(4) + y = 1 - color(red)(4)#

#0 + y = -3#

#y = -3# or #(2, -3)#

Equation 2:

For #y = 0#:

#x + (2 * 0) = -2#

#x + 0 = -2#

#x = -2# or #(-2, 0)#

For #x = 0#:

#0 + 2y = -2#

#2y = -2#

#(2y)/color(red)(2) = -2/color(red)(2)#

#y = -1# or #(0, -1)#

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]}

Or

#(8/3, -5/3)#