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

Question

4404 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-23T13:35:46

See a solution process below:

First, solve for two points on the first equation, plot the two points and then draw a straight line through the two points:

First Point: For #x = 0#

#y = 1/2 * 0#

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

Second Point: For #x = 2#

#y = 1/2 * 2#

#y = 1# or #(2, 1)#

graph{(y - 0.5x)(x^2+y^2-0.035)((x-2)^2+(y-1)^2-0.035)=0 [-10, 10, -5, 5]}

Next, solve for two points on the second equation, plot the two points and then draw a straight line through the two points:

First Point: For #x = 0#

#y = -0 + 3#

#y = 3# or #(0, 3)#

Second Point: For #x = 3#

#y = -3 + 3#

#y = 0# or #(3, 0)#

graph{(y+x-3)(y - 0.5x)(x^2+(y-3)^2-0.035)((x-3)^2+y^2-0.035)=0 [-10, 10, -5, 5]}

We can see the lines intersect at #(2, 1)#

graph{(y+x-3)(y - 0.5x)((x-2)^2+(y-1)^2-0.035)=0 [-10, 10, -5, 5]}