QuestionB) Graph the inequality.
Studdy Solution
STEP 1
What is this asking?
We need to draw the line for and then shade the region that represents .
Watch out!
Remember that when we multiply or divide both sides of an inequality by a negative number, we *must* flip the inequality sign!
Also, a dashed line means points on the line are *not* included in the solution.
STEP 2
1. Rewrite the inequality as a linear equation.
2. Find two points that satisfy the equation.
3. Graph the line using the two points, remembering to use a dashed line.
4. Choose a test point *not* on the line.
5. Check if the test point satisfies the original inequality.
6. Shade the correct region.
STEP 3
Let's **rewrite** our inequality as an *equation* to find the boundary line: .
This helps us visualize the line that separates the solution region from the non-solution region.
STEP 4
To graph the line, we need two points.
Let's pick .
Substituting this into our equation , we get , which simplifies to .
Dividing both sides by gives us .
So, our **first point** is .
STEP 5
Now, let's pick .
Substituting into , we get , which simplifies to .
Dividing both sides by gives us .
So, our **second point** is .
STEP 6
Plot the two points and on the coordinate plane.
Since our original inequality is (strictly less than), we draw a **dashed line** through these points.
The dashed line indicates that points *on* the line are *not* part of the solution.
STEP 7
A convenient test point is the **origin** because it makes calculations easy!
Make sure your test point isn't on the line we just drew.
STEP 8
Substitute the test point into the original inequality .
We get , which simplifies to .
This is **true**!
STEP 9
Since our test point satisfies the inequality, we shade the region that *contains* the origin.
This is the region **above** the dashed line.
If the test point had *not* satisfied the inequality, we would have shaded the *opposite* side of the line.
STEP 10
The solution is the region above the dashed line , including all points within that region but *not* the points on the line itself.
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