How do you solve #-x + 3y =10#, #x+y=2# by graphing and classify the system?

2 Answers

Answer:

Two unknowns and two equations #x=-1# and #y=3#

Explanation:

Add these equations first:
#-x+x+3y+y=10+2#
#4y=12#
#y=3#
If #y=3# what is x?
#-x+3*3=10#
#-x=10-9#
#x=-1#

The first equation (GRAPH) graph{(10+x)/3 [-10, 10, -5, 5]}

The second equation (GRAPH) graph{2-x [-10, 10, -5, 5]}

Together: graph{(-x+3y-10)(x+y-2)=0 [-10, 10, -5, 5]}

Apr 12, 2017

Answer:

Se below for solution by (only) graphing

Explanation:

Pick a pair of sample points for the equation #color(green)(-x+3y=10)#
#color(white)("XXX"){: (color(green)(ul(x)),color(green)(ul(y))), (color(green)(5),color(green)(5)), (color(green)(2),color(green)(4)) :}#

Pick a pair of sample points for the equation #color(magenta)(x+y=2)#
#color(white)("XXX"){: (color(magenta)(ul(x)),color(magenta)(ul(y))), (color(magenta)(0),color(magenta)(2)), (color(magenta)(2),color(magenta)(0)) :}#

Plot the points for each equation and draw a line through the pair for each:
enter image source here
Read the solution as the intersection point directly from the graph:
#color(white)("XXX")(x,y)=(-1,3)#