Question

How do you graph `y< x + 4` and `x- 2y = 10`?

Answer 1

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Please read the explanation.

Explanation:

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Given the Compound Inequality:

`color(red)((y< x + 4) and (x- 2y = 10)`

`y < x+4` ... Result.1

Next, consider

`x- 2y = 10`

We will bring it down to Slope-Intercept Form.

Subtract `color(red)(x)` from both sides"

`rArr x- 2y-x = 10-x`

`rArr cancel x- 2y-cancel x = 10-x`

`rArr -2y=10-x`

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

`rArr (-1)(-2y)=(-1)(10-x)`

`rArr 2y=-10+x`

Rearrange the terms:

`rArr 2y=x-10`

Divide both sides by `color(red)(2`

`rArr (2y)/2=(x-10)/2`

`rArr (cancel 2y)/cancel 2=(x-10)/2`

`rArr y=(1/2)*(x-10)` ... Result.2

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

We will create data tables using both the intermediate results:

The first `color(red)((x,y)` table is for the first inequality.

The second `color(red)((x,y)` table is for the equality.

Next, we can graph:

Dotted line are values NOT part of the solution.

Shaded area is the solution region for the inequality.

The second equation is a linear equation creates a straight line.

Since `color(blue)("AND")` is used in the problem, we need to choose all the values that are common for both.

This would mean, points lying on the straight line, for the equation below, will be the final solution.

`y=(1/2)*(x-10)` ... Result.2

Points on the straight line common for both the graphs is the final solution.

Observe that NOT ALL POINTS on the line belong to the solution:

Points beyond `color(red)((-18, -14)` on the negative side of x-axis are NOT part of the final solution.

Hope it helps.