Answer :
To solve this problem, we need to determine the largest possible value of [tex]\( y \)[/tex] given the two inequalities:
1. [tex]\( 8 \leq 4x \leq 20 \)[/tex]
2. [tex]\( y - 3x < 12 \)[/tex]
Let's break down the solution step-by-step:
### Step 1: Solve the first inequality for [tex]\( x \)[/tex].
The inequality is [tex]\( 8 \leq 4x \leq 20 \)[/tex]. We can solve this by dividing every part of the inequality by 4:
[tex]\[ \frac{8}{4} \leq x \leq \frac{20}{4} \][/tex]
This simplifies to:
[tex]\[ 2 \leq x \leq 5 \][/tex]
Since [tex]\( x \)[/tex] must be an integer, the possible values for [tex]\( x \)[/tex] are 2, 3, 4, and 5.
### Step 2: Use the values of [tex]\( x \)[/tex] to solve the second inequality for [tex]\( y \)[/tex].
The second inequality is [tex]\( y - 3x < 12 \)[/tex]. To find the largest possible value for [tex]\( y \)[/tex], we rearrange the inequality:
[tex]\[ y < 12 + 3x \][/tex]
We will determine [tex]\( y \)[/tex] for each integer value of [tex]\( x \)[/tex] identified in Step 1:
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ y < 12 + 3 \times 2 \][/tex]
[tex]\[ y < 12 + 6 \][/tex]
[tex]\[ y < 18 \][/tex]
The largest integer [tex]\( y \)[/tex] can be is 17.
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ y < 12 + 3 \times 3 \][/tex]
[tex]\[ y < 12 + 9 \][/tex]
[tex]\[ y < 21 \][/tex]
The largest integer [tex]\( y \)[/tex] can be is 20.
- For [tex]\( x = 4 \)[/tex]:
[tex]\[ y < 12 + 3 \times 4 \][/tex]
[tex]\[ y < 12 + 12 \][/tex]
[tex]\[ y < 24 \][/tex]
The largest integer [tex]\( y \)[/tex] can be is 23.
- For [tex]\( x = 5 \)[/tex]:
[tex]\[ y < 12 + 3 \times 5 \][/tex]
[tex]\[ y < 12 + 15 \][/tex]
[tex]\[ y < 27 \][/tex]
The largest integer [tex]\( y \)[/tex] can be is 26.
### Step 3: Determine the largest possible value of [tex]\( y \)[/tex].
From the possible solutions for each value of [tex]\( x \)[/tex], the largest possible value for [tex]\( y \)[/tex] is 26.
Therefore, the largest possible value of [tex]\( y \)[/tex] that satisfies both inequalities is 26.
1. [tex]\( 8 \leq 4x \leq 20 \)[/tex]
2. [tex]\( y - 3x < 12 \)[/tex]
Let's break down the solution step-by-step:
### Step 1: Solve the first inequality for [tex]\( x \)[/tex].
The inequality is [tex]\( 8 \leq 4x \leq 20 \)[/tex]. We can solve this by dividing every part of the inequality by 4:
[tex]\[ \frac{8}{4} \leq x \leq \frac{20}{4} \][/tex]
This simplifies to:
[tex]\[ 2 \leq x \leq 5 \][/tex]
Since [tex]\( x \)[/tex] must be an integer, the possible values for [tex]\( x \)[/tex] are 2, 3, 4, and 5.
### Step 2: Use the values of [tex]\( x \)[/tex] to solve the second inequality for [tex]\( y \)[/tex].
The second inequality is [tex]\( y - 3x < 12 \)[/tex]. To find the largest possible value for [tex]\( y \)[/tex], we rearrange the inequality:
[tex]\[ y < 12 + 3x \][/tex]
We will determine [tex]\( y \)[/tex] for each integer value of [tex]\( x \)[/tex] identified in Step 1:
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ y < 12 + 3 \times 2 \][/tex]
[tex]\[ y < 12 + 6 \][/tex]
[tex]\[ y < 18 \][/tex]
The largest integer [tex]\( y \)[/tex] can be is 17.
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ y < 12 + 3 \times 3 \][/tex]
[tex]\[ y < 12 + 9 \][/tex]
[tex]\[ y < 21 \][/tex]
The largest integer [tex]\( y \)[/tex] can be is 20.
- For [tex]\( x = 4 \)[/tex]:
[tex]\[ y < 12 + 3 \times 4 \][/tex]
[tex]\[ y < 12 + 12 \][/tex]
[tex]\[ y < 24 \][/tex]
The largest integer [tex]\( y \)[/tex] can be is 23.
- For [tex]\( x = 5 \)[/tex]:
[tex]\[ y < 12 + 3 \times 5 \][/tex]
[tex]\[ y < 12 + 15 \][/tex]
[tex]\[ y < 27 \][/tex]
The largest integer [tex]\( y \)[/tex] can be is 26.
### Step 3: Determine the largest possible value of [tex]\( y \)[/tex].
From the possible solutions for each value of [tex]\( x \)[/tex], the largest possible value for [tex]\( y \)[/tex] is 26.
Therefore, the largest possible value of [tex]\( y \)[/tex] that satisfies both inequalities is 26.
