How do you find the slope and intercept of #x-y=1#?

This equation is in Standard Linear form. The standard form of a linear equation is: #color(red)(A)x + color(blue)(B)y = color(green)(C)#

Where, if at all possible, #color(red)(A)#, #color(blue)(B)#, and #color(green)(C)#are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

The slope of an equation in standard form is: #m = -color(red)(A)/color(blue)(B)#

The #y#-intercept of an equation in standard form is: #color(green)(C)/color(blue)(B)#

#color(red)(1)x + color(blue)(-1)y = color(green)(1)#

Therefore:

• The slope is: #m = (-color(red)(1))/color(blue)(-1) = 1#

• The #y#-intercept is: #color(green)(1)/color(blue)(-1) = -1# or #(0, -1)#

Given equation of straight line is

#x-y=1#

#x/1+y/-1=1#

The above equation is in standard intercept form of line: #x/a+y/b=1# which has

x-intercept: #a=1#

y-intercept: #b=-1#

The given equation of line:

#x-y=1#

#y=x-1#

The above equation is in standard slope-intercept form: #y=mx+c# with slope

#m=1 #

Slope: #m=1#

Recall slope-intercept form

#y=mx+b#, with slope #m# and a #y#-intercept of #b#.

We essentially just want a #y# on the left side. Let's subtract #x# from both sides to get

#-y=-x+1#

Next, divide both sides by #-1# to get

#y=x-1#

Now, our equation is in slope-intercept form, with a slope of #1#, and a #y#-intercept of #-1#.

Hope this helps!