How do you graph `x-y=1`?

Graph:

`x-y=1`

X-intercept: value of `x` when `y=0`

Substitute `0` for `y` and solve for `x`.

x-0=1#

`x=1`

The x-intercept is `(1,0)`. Plot this point.

Y-intercept: value of `y` when `x=0`

Substitute `0` for `x` and solve for `y`.

`0-y=1`

`-y=1`

Multiply both sides by `-1`. This will reverse the signs.

`-1(-y)=1xx-1`

`y=-1`

The y-intercept is `(0,-1)`. Plot this point.

Draw a straight line through the points.

graph{x-y=1 [-10, 10, -5, 5]}

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We are given the equation: `color(red)(x-y=1`

How do we graph this linear equation ?

We have

`color(blue)(x-y=1`

The given equation can be reduced to a more simpler form.

Subtract `color(red)(x` from both sides of the equation:

`rArr x-y-x=1-x`

`rArr cancel x-y-cancel x=1-x`

`rArr -y = 1-x`

Multiply both sides of the equation by `color(red)((-1)`

`rArr -y*(-1) = (1-x)(-1)`

`rArr y=x-1`

Now, we have the equation in Slope-Intercept Form:

`color(blue)(y=mx+b)`, where

`color(red)(m` is the Slope and

`color(red)(b` is the y-intercept.

Hence,

Slope : `color(blue)(1`

y-intercept :`color(blue)((0,-1)`

The x-intercept is where the graph crosses the x-axis

Set `color(red)(y=0` to find the x-intercept:

`rArr x-1=0`

Add `color(red)(1` to both sides of the equation:

`rArr x-1+1=0+1`

`rArr x-cancel 1+cancel 1=0+1`

`rArr x=1`

Hence `color(red)((1,0)` is the x-intercept.

Plot the points x-intercept and intercept on the graph.

Join the two points to get the straight line.

This is the graph of the linear equation.

Hope you find this solution useful.