How do you solve the system by graphing #y= 1/4x -1# and #y= -2x -10#?

2 Answers

Answer:

#x=-4#
#y=-2#

Explanation:

the given system

#y=1/4x-1#

#y=-2x-10#

Solution

#y=y#

#1/4x-1=-2x-10#

Transpose all terms with x on the left side

#1/4x+2x=-10+1#

#(1/4+2)x=-9#

#9/4x=-9#

Multiply now both sides by 4/9

#4/9*9/4x=4/9(-9)#

#x=-4#

to solve for #y# use equation #y=-2x-10# and #x=-4#

#y=-2x-10#

#y=-2(-4)-10#

#y=8-10#

#y=-2#

Checking

With equation #y=1/4x-1#

#y=1/4x-1#

#-2=1/4(-4)-1#

#-2=-2" "#correct

~~~~~~~~~~~~~~~~~~~~~~~

Checking

With equation #y=-2x-10#

#y=-2x-10#

#-2=-2(-4)-10#

#-2=-2" "#correct

God bless....I hope the explanation is useful

Jun 20, 2016

Answer:

Solution by graphing below.
#color(white)("XXX")(x,y)~~(-4,-2)#

Explanation:

For each of the given equations select a few (at least two but I would suggest three) values for #x# and solve for corresponding #y# values.

#{: (color(red)(y=1/4x-1),,," | ",color(blue)(y=-2x-10),,), (,color(red)(underline(x)),color(white)("X")color(red)(underline(y))," | ",,color(white)("X")color(blue)(underline(x)),color(white)("X")color(blue)(underline(y))), (,color(red)(0),color(red)(-1)," | ",,color(white)("X")color(blue)(0),color(blue)(-10)), (,color(red)(4),color(white)("X")color(red)(0)," | ",,color(white)("X")color(blue)(1),color(blue)(-12)), (,color(red)(8),color(white)("X")color(red)(1)," | ",,color(blue)(-1),color(white)("X")color(blue)(-8)) :}#

Plot each pair of points which you have calculated and draw a line for each equation's set of points.
enter image source here
The point where the two equation lines intersect provide the coordinates of the system solution.

In this case it appears that #x# is about #-4# and #y# is about #-2#.
(This in fact the exact solution but, in general, solving equations by graphing is very dependent on the accuracy of your drawing).