Math  /  Geometry

Questiony>x4y14x+1\begin{array}{l} y>-x-4 \\ y \leq \frac{1}{4} x+1 \end{array}
Answer

Studdy Solution

STEP 1

What is this asking? We need to graph two inequalities on the same coordinate plane and find the overlapping shaded region! Watch out! Don't forget to check if the lines themselves are included in the solution, based on whether the inequality is "less than or equal to" or just "less than".

STEP 2

1. Graph the first inequality
2. Graph the second inequality
3. Find the overlapping region

STEP 3

Alright, let's **start** with the first inequality: y>x4y > -x - 4.
Imagine it as an equation for a moment: y=x4y = -x - 4.
This is a **line**!
Its **slope** is 1-1 and its **y-intercept** is 4-4.

STEP 4

To **graph** this line, we can start by plotting the **y-intercept**, which is the point (0,4)(0, -4).
From there, we use the **slope**!
A slope of 1-1 means that for every **1 unit** we move to the right along the x-axis, we move **1 unit down** along the y-axis.
So, from (0,4)(0, -4), we can plot another point at (1,5)(1, -5).

STEP 5

Now, **draw a dashed line** through these points.
It's *dashed* because our inequality is y>x4y > -x - 4, *not* yx4y \geq -x - 4, meaning the line itself isn't included in the solution.

STEP 6

Time to **shade**!
Since the inequality is y>x4y > -x - 4, we want to shade the area *above* the line.
Pick a **test point** not on the line, like (0,0)(0, 0).
Is 0>040 > -0 - 4?
Yes! 0>40 > -4.
So, we shade the side of the line that includes (0,0)(0, 0).

STEP 7

On to the second inequality: y14x+1y \leq \frac{1}{4}x + 1.
Again, think of it as an equation: y=14x+1y = \frac{1}{4}x + 1.
Here, the **slope** is 14\frac{1}{4} and the **y-intercept** is 11.

STEP 8

Plot the **y-intercept** at (0,1)(0, 1).
The **slope** of 14\frac{1}{4} means that for every **4 units** we move right, we move **1 unit up**.
So, from (0,1)(0, 1), we can plot another point at (4,2)(4, 2).

STEP 9

Now, **draw a solid line** through these points.
It's *solid* because our inequality is y14x+1y \leq \frac{1}{4}x + 1, meaning the line itself *is* included in the solution.

STEP 10

For **shading**, since it's y14x+1y \leq \frac{1}{4}x + 1, we shade *below* the line.
Using (0,0)(0, 0) as a **test point** again: Is 014(0)+10 \leq \frac{1}{4}(0) + 1?
Yes! 010 \leq 1.
So, we shade the side of the line that includes (0,0)(0, 0).

STEP 11

The **solution** to the system of inequalities is the region where the shading from *both* inequalities overlaps!
This is the region that satisfies *both* y>x4y > -x - 4 and y14x+1y \leq \frac{1}{4}x + 1 simultaneously.

STEP 12

The solution is the overlapping shaded region on the graph, including the solid line from y14x+1y \leq \frac{1}{4}x + 1 within the overlapping region, but *excluding* the dashed line from y>x4y > -x - 4.

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