Math / Geometry

QuestionSolve the following system of inequalities graphically on the set of axes below. State th coordinates of a point in the solution set. $\begin{array}{c} y>3 x-4 \\ y \geq-\frac{1}{2} x+3 \end{array}$
Line 2 Change line Change shade

Studdy Solution

STEP 1

What is this asking? We need to find all the $(x, y)$ points where $y$ is greater than $3x - 4$ and $y$ is greater than or equal to $-\frac{1}{2}x + 3$ !
We'll graph these inequalities and see where the shaded regions overlap. Watch out! Remember, a dashed line means the points on the line aren't included in the solution, while a solid line means they are!

STEP 2

1. Graph the first inequality
2. Graph the second inequality
3. Identify the solution area

STEP 3

Let's graph $y > 3x - 4$ .
We'll start by graphing the line $y = 3x - 4$ .
This line has a slope of $3$ (meaning it goes up $3$ units for every $1$ unit to the right) and a y-intercept of $-4$ (meaning it crosses the y-axis at $-4$ ).

STEP 4

Since the inequality is $y > 3x - 4$ (strictly greater than, not "greater than or equal to"), we'll draw a dashed line to show that the points on the line itself aren't part of the solution.

STEP 5

Now, we need to shade the region above the line because we want the $y$ values that are greater than $3x - 4$ .
Imagine plugging in $(0, 0)$ .
Is $0 > 3 \cdot 0 - 4$ ?
Yes, $0 > -4$ , so the side with $(0, 0)$ is the correct side to shade!

STEP 6

Now let's graph $y \geq -\frac{1}{2}x + 3$ .
We'll graph the line $y = -\frac{1}{2}x + 3$ .
This line has a slope of $-\frac{1}{2}$ (meaning it goes down $1$ unit for every $2$ units to the right) and a y-intercept of $3$ .

STEP 7

Since the inequality is $y \geq -\frac{1}{2}x + 3$ (greater than or equal to), we draw a solid line because the points on the line are included in the solution.

STEP 8

We want $y$ values greater than or equal to $-\frac{1}{2}x + 3$ , so we shade the region above the line.
Again, imagine plugging in $(0, 0)$ .
Is $0 \geq -\frac{1}{2} \cdot 0 + 3$ ?
No, $0$ is not greater than or equal to $3$ , so we shade the other side, the one that doesn't include $(0, 0)$ .

STEP 9

The solution to the system of inequalities is the region where the shading from both inequalities overlaps!
This is the region that satisfies both conditions at the same time.
#### 2.3.2 We need to pick a point in the solution set.
Looking at the graph, $(2, 4)$ seems to be in the overlapping shaded region.
Let's check: Is $4 > 3 \cdot 2 - 4$ ?
Yes, $4 > 2$ . Is $4 \geq -\frac{1}{2} \cdot 2 + 3$ ?
Yes, $4 \geq 2$ . So, $(2, 4)$ is definitely a solution!

STEP 10

The solution is the overlapping shaded region on the graph.
One point in the solution set is $(2, 4)$ .

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QuestionSolve the following system of inequalities graphically on the set of axes below. State th coordinates of a point in the solution set. $\begin{array}{c} y>3 x-4 \\ y \geq-\frac{1}{2} x+3 \end{array}$
Line 2 Change line Change shade

Studdy Solution

STEP 1

STEP 2

1. Graph the first inequality
2. Graph the second inequality
3. Identify the solution area

STEP 3

Let's graph $y > 3x - 4$ .
We'll start by graphing the line $y = 3x - 4$ .
This line has a slope of $3$ (meaning it goes up $3$ units for every $1$ unit to the right) and a y-intercept of $-4$ (meaning it crosses the y-axis at $-4$ ).

STEP 4

Since the inequality is $y > 3x - 4$ (strictly greater than, not "greater than or equal to"), we'll draw a dashed line to show that the points on the line itself aren't part of the solution.

STEP 5

Now, we need to shade the region above the line because we want the $y$ values that are greater than $3x - 4$ .
Imagine plugging in $(0, 0)$ .
Is $0 > 3 \cdot 0 - 4$ ?
Yes, $0 > -4$ , so the side with $(0, 0)$ is the correct side to shade!

STEP 6

Now let's graph $y \geq -\frac{1}{2}x + 3$ .
We'll graph the line $y = -\frac{1}{2}x + 3$ .
This line has a slope of $-\frac{1}{2}$ (meaning it goes down $1$ unit for every $2$ units to the right) and a y-intercept of $3$ .

STEP 7

Since the inequality is $y \geq -\frac{1}{2}x + 3$ (greater than or equal to), we draw a solid line because the points on the line are included in the solution.

STEP 8

We want $y$ values greater than or equal to $-\frac{1}{2}x + 3$ , so we shade the region above the line.
Again, imagine plugging in $(0, 0)$ .
Is $0 \geq -\frac{1}{2} \cdot 0 + 3$ ?
No, $0$ is not greater than or equal to $3$ , so we shade the other side, the one that doesn't include $(0, 0)$ .

STEP 9

STEP 10

The solution is the overlapping shaded region on the graph.
One point in the solution set is $(2, 4)$ .

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QuestionSolve the following system of inequalities graphically on the set of axes below. State th coordinates of a point in the solution set. y>3x−4y≥−12x+3\begin{array}{c} y>3 x-4 \\ y \geq-\frac{1}{2} x+3 \end{array}y>3x−4y≥−21​x+3​Line 2 Change line Change shade

Studdy Solution

STEP 1

STEP 2

1. Graph the first inequality2. Graph the second inequality3. Identify the solution area

STEP 3

Let's **graph** y>3x−4y > 3x - 4y>3x−4.We'll start by graphing the line y=3x−4y = 3x - 4y=3x−4.This line has a **slope** of 333 (meaning it goes up 333 units for every 111 unit to the right) and a **y-intercept** of −4-4−4 (meaning it crosses the y-axis at −4-4−4).

STEP 4

Since the inequality is y>3x−4y > 3x - 4y>3x−4 (strictly greater than, not "greater than or equal to"), we'll draw a **dashed line** to show that the points on the line itself *aren't* part of the solution.

STEP 5

Now, we need to **shade** the region above the line because we want the yyy values that are *greater* than 3x−43x - 43x−4.Imagine plugging in (0,0)(0, 0)(0,0).Is 0>3⋅0−40 > 3 \cdot 0 - 40>3⋅0−4?Yes, 0>−40 > -40>−4, so the side with (0,0)(0, 0)(0,0) is the correct side to shade!

STEP 6

Now let's **graph** y≥−12x+3y \geq -\frac{1}{2}x + 3y≥−21​x+3.We'll graph the line y=−12x+3y = -\frac{1}{2}x + 3y=−21​x+3.This line has a **slope** of −12-\frac{1}{2}−21​ (meaning it goes *down* 111 unit for every 222 units to the right) and a **y-intercept** of 333.

STEP 7

Since the inequality is y≥−12x+3y \geq -\frac{1}{2}x + 3y≥−21​x+3 (greater than *or equal to*), we draw a **solid line** because the points on the line *are* included in the solution.

STEP 8

STEP 9

STEP 10

The solution is the overlapping shaded region on the graph.One point in the solution set is (2,4)(2, 4)(2,4).

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QuestionSolve the following system of inequalities graphically on the set of axes below. State th coordinates of a point in the solution set. $\begin{array}{c} y>3 x-4 \\ y \geq-\frac{1}{2} x+3 \end{array}$
Line 2 Change line Change shade

1. Graph the first inequality
2. Graph the second inequality
3. Identify the solution area

Let's graph $y > 3x - 4$ .
We'll start by graphing the line $y = 3x - 4$ .
This line has a slope of $3$ (meaning it goes up $3$ units for every $1$ unit to the right) and a y-intercept of $-4$ (meaning it crosses the y-axis at $-4$ ).

Since the inequality is $y > 3x - 4$ (strictly greater than, not "greater than or equal to"), we'll draw a dashed line to show that the points on the line itself aren't part of the solution.

Now, we need to shade the region above the line because we want the $y$ values that are greater than $3x - 4$ .
Imagine plugging in $(0, 0)$ .
Is $0 > 3 \cdot 0 - 4$ ?
Yes, $0 > -4$ , so the side with $(0, 0)$ is the correct side to shade!

Now let's graph $y \geq -\frac{1}{2}x + 3$ .
We'll graph the line $y = -\frac{1}{2}x + 3$ .
This line has a slope of $-\frac{1}{2}$ (meaning it goes down $1$ unit for every $2$ units to the right) and a y-intercept of $3$ .

Since the inequality is $y \geq -\frac{1}{2}x + 3$ (greater than or equal to), we draw a solid line because the points on the line are included in the solution.

The solution is the overlapping shaded region on the graph.
One point in the solution set is $(2, 4)$ .