How do you graph #x+5y=5# using intercepts?

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Answer 1 · 2018-07-03T04:43:56

Given function

#x+5y=5#

#\frac{x}{5}+\frac{5y}{5}=1#

#x/5+y/1=1#

The above line has x-intercept #=5#

y-intercept #=1#

i,e, the line intersects the coordinate axes at the points #(5, 0)# & #(0, 1)#

Now, specify the points #(5, 0)# & #(5, 0)# & join these points to get plot of given straight line

Answer 2 · 2018-07-03T04:45:27

Please see below.

First find the #x#-intercept by putting #y=0# i.e. #x=5#, which gives us point #(5,0)#

then find #y#-intercept by putting #x=0# i.e. #5y=5# or #y=5/5=1#, giving us another point #(0,1)#.

Joining two points gives us the graph of #x+5y=5#, as shown below

graph{(x+5y-5)(x^2+(y-1)^2-0.01)((x-5)^2+y^2-0.01)=0 [-3.42, 6.58, -1.42, 3.58]}

Answer 3 · 2018-07-03T04:46:37

Use two arbitrary values for #x#. For example, find the intercept with #y# if #x=0#.

#0+5y=5#
#5ycolor(red)(/5)=5color(red)(/5)#
#y=1#

Now you know one of the intercepts is #(0,1)#. Continue using another random #x# value. Let's try #x=5#.

#cancel(5color(red)(-5))+5y=5color(red)(-5)#
#5ycolor(red)(/5)=0color(red)(/5)#
#y=0#

The second intercept is #(5,0)#.

Using these two intercepts, you can map out an x-axis and y-axis on some graph paper; draw these two intercepts and connect a line through them.

Answer 4 · 2018-07-03T04:48:58

As below

#x + 5y = 5#

#5y = -x + 5#

#y = -x/5 + 1#

Y-intercept = 1#

X-intercept when y = 0

#x/5 = 1#

X-intercept = 5#

Two points are (5, 0), (0, 1)

graph{-x/5 + 1 [-10, 10, -5, 5]}