Answer :
To solve this problem, we need to represent the given situation with a system of equations and then solve for the length and width of the rectangle.
### Problem Breakdown:
1. Understanding the Rectangle's Dimensions:
- You are given that the length ([tex]\(l\)[/tex]) of a rectangle is 4 units longer than twice its width ([tex]\(w\)[/tex]).
- This relationship can be written as an equation:
[tex]\[
l = 2w + 4
\][/tex]
2. Rectangle's Area:
- The area of a rectangle is given by multiplying its length and width, and you are told that this area is 126 square units.
- Therefore, you can form the second equation:
[tex]\[
w \times l = 126
\][/tex]
### System of Equations:
Combining these two pieces of information, the system of equations that represents this situation is:
- [tex]\(l = 2w + 4\)[/tex]
- [tex]\(w \times l = 126\)[/tex]
### Solving the System:
1. Plug the Expression for [tex]\(l\)[/tex] into the Area Equation:
- Substitute [tex]\(l = 2w + 4\)[/tex] into the area equation [tex]\(w \times l = 126\)[/tex]:
[tex]\[
w \times (2w + 4) = 126
\][/tex]
2. Expand and Simplify:
- Multiply [tex]\(w\)[/tex] through the parentheses:
[tex]\[
2w^2 + 4w = 126
\][/tex]
- Bring all terms to one side to form a quadratic equation:
[tex]\[
2w^2 + 4w - 126 = 0
\][/tex]
3. Solve the Quadratic Equation:
- You can simplify by dividing the entire equation by 2:
[tex]\[
w^2 + 2w - 63 = 0
\][/tex]
- Factor the quadratic equation:
[tex]\[
(w - 7)(w + 9) = 0
\][/tex]
- Solving for [tex]\(w\)[/tex], we find:
[tex]\[
w = 7 \quad \text{or} \quad w = -9
\][/tex]
- Since width cannot be negative, [tex]\(w = 7\)[/tex].
4. Find the Length:
- Use the equation [tex]\(l = 2w + 4\)[/tex] with [tex]\(w = 7\)[/tex]:
[tex]\[
l = 2(7) + 4 = 14 + 4 = 18
\][/tex]
### Conclusion:
The width of the rectangle is 7 units, and the length is 18 units. The correct answer choice that matches this system of equations is option d: [tex]\(l = 2w + 4\)[/tex] and [tex]\(w \cdot l = 126\)[/tex].
### Problem Breakdown:
1. Understanding the Rectangle's Dimensions:
- You are given that the length ([tex]\(l\)[/tex]) of a rectangle is 4 units longer than twice its width ([tex]\(w\)[/tex]).
- This relationship can be written as an equation:
[tex]\[
l = 2w + 4
\][/tex]
2. Rectangle's Area:
- The area of a rectangle is given by multiplying its length and width, and you are told that this area is 126 square units.
- Therefore, you can form the second equation:
[tex]\[
w \times l = 126
\][/tex]
### System of Equations:
Combining these two pieces of information, the system of equations that represents this situation is:
- [tex]\(l = 2w + 4\)[/tex]
- [tex]\(w \times l = 126\)[/tex]
### Solving the System:
1. Plug the Expression for [tex]\(l\)[/tex] into the Area Equation:
- Substitute [tex]\(l = 2w + 4\)[/tex] into the area equation [tex]\(w \times l = 126\)[/tex]:
[tex]\[
w \times (2w + 4) = 126
\][/tex]
2. Expand and Simplify:
- Multiply [tex]\(w\)[/tex] through the parentheses:
[tex]\[
2w^2 + 4w = 126
\][/tex]
- Bring all terms to one side to form a quadratic equation:
[tex]\[
2w^2 + 4w - 126 = 0
\][/tex]
3. Solve the Quadratic Equation:
- You can simplify by dividing the entire equation by 2:
[tex]\[
w^2 + 2w - 63 = 0
\][/tex]
- Factor the quadratic equation:
[tex]\[
(w - 7)(w + 9) = 0
\][/tex]
- Solving for [tex]\(w\)[/tex], we find:
[tex]\[
w = 7 \quad \text{or} \quad w = -9
\][/tex]
- Since width cannot be negative, [tex]\(w = 7\)[/tex].
4. Find the Length:
- Use the equation [tex]\(l = 2w + 4\)[/tex] with [tex]\(w = 7\)[/tex]:
[tex]\[
l = 2(7) + 4 = 14 + 4 = 18
\][/tex]
### Conclusion:
The width of the rectangle is 7 units, and the length is 18 units. The correct answer choice that matches this system of equations is option d: [tex]\(l = 2w + 4\)[/tex] and [tex]\(w \cdot l = 126\)[/tex].
