Question

Solve for x:
9x + 16 = 6x + 1
4x - 3 = 3x - 9
13x + 26 = 3x - 34


Solve ax:
ax - 12 = 9


The area of a rectangle is calculated with the formula A = lw. If the area of a rectangle is 52.5m ^2 and the width is 5 m, what is the length? Explain the solution to the real-world problem.(1 point)

Answer 1

Let’s start solving the equations one by one.

### First Set of Equations

1. **Equation 1:**
\[
9x + 16 = 6x + 1
\]
To isolate \( x \), first subtract \( 6x \) from both sides:
\[
9x - 6x + 16 = 1
\]
This simplifies to:
\[
3x + 16 = 1
\]
Next, subtract 16 from both sides:
\[
3x = 1 - 16
\]
Simplifying further:
\[
3x = -15
\]
Finally, divide by 3:
\[
x = -5
\]

2. **Equation 2:**
\[
4x - 3 = 3x - 9
\]
To isolate \( x \), first subtract \( 3x \) from both sides:
\[
4x - 3x - 3 = -9
\]
This simplifies to:
\[
x - 3 = -9
\]
Next, add 3 to both sides:
\[
x = -9 + 3
\]
Thus, we find:
\[
x = -6
\]

3. **Equation 3:**
\[
13x + 26 = 3x - 34
\]
First, subtract \( 3x \) from both sides:
\[
13x - 3x + 26 = -34
\]
Which simplifies to:
\[
10x + 26 = -34
\]
Next, subtract 26 from both sides:
\[
10x = -34 - 26
\]
Simplifying gives:
\[
10x = -60
\]
Divide by 10:
\[
x = -6
\]

### Summary of the First Set:
- From the three equations we solved, we got:
- \( x = -5 \)
- \( x = -6 \)
- \( x = -6 \)

### Second Set of Equation

**Solve for \( ax \):**
\[
ax - 12 = 9
\]
First, add 12 to both sides:
\[
ax = 9 + 12
\]
Simplifying gives:
\[
ax = 21
\]

### Real-World Problem

**Finding the Length of a Rectangle:**
Given the area \( A = lw \):
\[
A = 52.5 \text{ m}^2 \quad \text{and} \quad w = 5 \text{ m}
\]
Substituting these into the area formula gives:
\[
52.5 = l \times 5
\]
To find \( l \), divide both sides by 5:
\[
l = \frac{52.5}{5}
\]
Calculating this gives:
\[
l = 10.5 \text{ m}
\]

### Explanation:
In this real-world problem, we found that the area of the rectangle is 52.5 square meters and the width is 5 meters. We set up the formula for area \( A = lw \) and rearranged the formula to solve for length \( l \). This process is practical when designing products or spaces, where knowing dimensions is crucial for fitting spaces or ensuring resources meet requirements. Here, the rectangle with a width of 5 m will have a length of 10.5 m to maintain an area of 52.5 m².