Question

graphing systems of inequalities independent practice
in 1-3, graph the system of inequalities. then determine if each ordered pair represents a solution to the system.
1
$y < 5$
$y \geq \frac{1}{3}x + 3$

a. $(-3, 4)$
b. $(2, 5)$

2
$y \geq -\frac{1}{2}x$
$2x + 4y < 20$

a. $(2, 4)$
b. $(-3, 5)$

3
$y > 2x - 4$
$6x + 3y < 9$

a. $(0, 0)$
b. $(-3, -7)$
name: destiny cuevas
date: 12/8/25
pa 1

Explanation:

Step1: Analyze Problem 1a

We have the system of inequalities \( y < 5 \) and \( y\geq\frac{1}{3}x + 3 \). We need to check if the ordered pair \((-3,4)\) is a solution.
First, check \( y < 5 \): Substitute \( y = 4 \), we get \( 4<5 \), which is true.
Second, check \( y\geq\frac{1}{3}x + 3 \): Substitute \( x=-3 \) and \( y = 4 \).
Calculate \( \frac{1}{3}(-3)+3=-1 + 3=2 \).
Now, check if \( 4\geq2 \), which is true.
Since both inequalities are satisfied, \((-3,4)\) is a solution.

Step2: Analyze Problem 1b

For the ordered pair \((2,5)\) and the system \( y < 5 \) and \( y\geq\frac{1}{3}x + 3 \).
First, check \( y < 5 \): Substitute \( y = 5 \), we get \( 5<5 \), which is false.
So, \((2,5)\) is not a solution.

Step3: Analyze Problem 2a

The system is \( y\geq-\frac{1}{2}x \) and \( 2x + 4y<20 \).
Check the ordered pair \((2,4)\).
First, check \( y\geq-\frac{1}{2}x \): Substitute \( x = 2 \), \( y = 4 \).
Calculate \( -\frac{1}{2}(2)=-1 \). Check if \( 4\geq - 1 \), which is true.
Second, check \( 2x + 4y<20 \): Substitute \( x = 2 \), \( y = 4 \).
Calculate \( 2(2)+4(4)=4 + 16=20 \). Check if \( 20<20 \), which is false.
So, \((2,4)\) is not a solution.

Step4: Analyze Problem 2b

For the ordered pair \((-3,5)\) and the system \( y\geq-\frac{1}{2}x \) and \( 2x + 4y<20 \).
First, check \( y\geq-\frac{1}{2}x \): Substitute \( x=-3 \), \( y = 5 \).
Calculate \( -\frac{1}{2}(-3)=\frac{3}{2}=1.5 \). Check if \( 5\geq1.5 \), which is true.
Second, check \( 2x + 4y<20 \): Substitute \( x=-3 \), \( y = 5 \).
Calculate \( 2(-3)+4(5)=-6 + 20 = 14 \). Check if \( 14<20 \), which is true.
Since both inequalities are satisfied, \((-3,5)\) is a solution.

Step5: Analyze Problem 3a

The system is \( y>2x - 4 \) and \( 6x+3y<9 \) (assuming the last inequality is \( 6x + 3y<9 \) as it's partially visible).
Check the ordered pair \((0,0)\).
First, check \( y>2x - 4 \): Substitute \( x = 0 \), \( y = 0 \).
We get \( 0>2(0)-4=-4 \), which is true.
Second, check \( 6x + 3y<9 \): Substitute \( x = 0 \), \( y = 0 \).
We get \( 6(0)+3(0)=0<9 \), which is true.
So, \((0,0)\) is a solution.

Step6: Analyze Problem 3b

For the ordered pair \((-3,-7)\) and the system \( y>2x - 4 \) and \( 6x+3y<9 \).
First, check \( y>2x - 4 \): Substitute \( x=-3 \), \( y=-7 \).
Calculate \( 2(-3)-4=-6 - 4=-10 \).
Check if \( -7>-10 \), which is true.
Second, check \( 6x + 3y<9 \): Substitute \( x=-3 \), \( y=-7 \).
Calculate \( 6(-3)+3(-7)=-18-21=-39 \).
Check if \( -39<9 \), which is true.
So, \((-3,-7)\) is a solution.

Answer:

1a. Yes (since \((-3,4)\) satisfies both inequalities)
1b. No (since \( y = 5 \) does not satisfy \( y<5 \))
2a. No (since \( 2(2)+4(4)=20 \) does not satisfy \( 2x + 4y<20 \))
2b. Yes (since \((-3,5)\) satisfies both inequalities)
3a. Yes (since \((0,0)\) satisfies both inequalities)
3b. Yes (since \((-3,-7)\) satisfies both inequalities)