Steply

Graph y <= -1/2x + 2 and y > x - 3

Step by step, with the reason for every step · Systems of inequalities · Algebra 1

Step by step

The question

Question: graph y ≤ −(1/2)x + 2 and y > x − 3.

① y ≤ −12x + 2 and ② y > x − 3

We shade the answer of each one.

Where both are shaded, both are true.

answer = the part shaded twice

1 Graph ①

Draw the grid.

①: The sign has "or equal to".

So the line is solid.

① ≤ → solid line

b = 2.

Start on the y-axis at 2.

m = −1/2.

Go right 2,

then down 1.

Dot.

Join the dots: solid.

Which side?

We test one point: (0, 0).

Put x = 0 and y = 0 into ①.

① 0 ≤ −12(0) + 2
0 ≤ 2 true

Is 0 less than or equal to 2?

Yes, true.

So we shade the side with (0, 0), in red.

2 Graph ②

②: The sign is more than, with no "or equal to".

So the line is dashed.

② > → dashed line

b = −3.

Start on the y-axis at −3.

m = 1.

Go right 1,

then up 1.

Dot.

Join the dots: dashed.

Which side?

We test one point: (0, 0).

Put x = 0 and y = 0 into ②.

② 0 > 0 − 3
0 > −3 true

Is 0 more than −3?

Yes, true.

So we shade the side with (0, 0), in blue.

3 Where both are shaded

This part is shaded twice.

It is the answer.

Careful!

The answer is not everything we shaded.

Only the part shaded twice works in both.

Careful: not every shaded part ✗

4 Check

Let's check with a point in the purple part: (−1, −1).

Put it into ①.

① −1 ≤ −12(−1) + 2
−1 ≤ 52 true

Now ②.

② −1 > (−1) − 3
−1 > −4 true

True in both!

So the answer is right.

Let's look back.

First: graph ①.

Then: graph ②.

Then: where both are shaded.

Last: check.

Answer
0 > −3 true

Your turn

Try this one yourself, then check:

y ≤ −12x + 2 and y > x − 4
Show the answer
the part shaded twice

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