5x-2y+1=6x+y, 3x+y+6=4x-y find area and perimeter of rectangle with respective sides

Let's solve the system of equations first to find the values of \(x\) and \(y\). The two equations are:

1. \( 5x - 2y + 1 = 6x + y \)
2. \( 3x + y + 6 = 4x - y \)

### Step 1: Simplify the first equation
Start by simplifying and solving for \(y\) in terms of \(x\).

\[
5x - 2y + 1 = 6x + y
\]

Subtract \(6x\) from both sides:

\[
5x - 6x - 2y + 1 = y
\]

\[
-x - 2y + 1 = y
\]

Now, subtract \(y\) from both sides:

\[
-x - 2y - y + 1 = 0
\]

\[
-x - 3y + 1 = 0
\]

Finally, move 1 to the other side:

\[
-x - 3y = -1
\]

This simplifies to:

\[
x + 3y = 1 \tag{3}
\]

### Step 2: Simplify the second equation
Now, simplify the second equation.

\[
3x + y + 6 = 4x - y
\]

Subtract \(4x\) from both sides:

\[
3x - 4x + y + 6 = -y
\]

\[
-x + y + 6 = -y
\]

Now, add \(y\) to both sides:

\[
-x + 2y + 6 = 0
\]

Finally, subtract 6 from both sides:

\[
-x + 2y = -6
\]

This simplifies to:

\[
x - 2y = 6 \tag{4}
\]

### Step 3: Solve the system of equations
Now, solve the system of equations formed by (3) and (4):

1. \( x + 3y = 1 \)
2. \( x - 2y = 6 \)

Subtract the second equation from the first:

\[
(x + 3y) - (x - 2y) = 1 - 6
\]

\[
x + 3y - x + 2y = -5
\]

This simplifies to:

\[
5y = -5
\]

\[
y = -1
\]

### Step 4: Substitute \(y = -1\) into one of the original equations
Substitute \(y = -1\) into \(x - 2y = 6\):

\[
x - 2(-1) = 6
\]

\[
x + 2 = 6
\]

\[
x = 4
\]

### Step 5: Find the area and perimeter of the rectangle
Now that we have \(x = 4\) and \(y = -1\), the dimensions of the rectangle are \(x = 4\) and \(y = 1\) (since dimensions are positive, we take the absolute value of \(y\)).

- **Area** = \(x \times y = 4 \times 1 = 4\)
- **Perimeter** = \(2(x + y) = 2(4 + 1) = 2(5) = 10\)

### Final Answer:
- Area of the rectangle = 4 square units
- Perimeter of the rectangle = 10 units

---

Would you like more details on any step, or do you have any questions?

Here are five related questions you might find interesting:
1. How would the solution change if \(y\) were negative in a different geometric context?
2. Can you solve a system of equations where both sides are quadratic?
3. What happens if you try solving these equations graphically?
4. How do you compute the area and perimeter for other shapes like triangles or circles?
5. Can the values of \(x\) and \(y\) represent something other than sides of a rectangle?

**Tip:** When solving systems of equations, always simplify as much as possible before trying to solve.

Learn how to solve the system of equations 5x - 2y + 1 = 6x + y and 3x + y + 6 = 4x - y to determine the dimensions of a rectangle. This problem covers algebraic concepts and geometry formulas such as area and perimeter of a rectangle. Suitable for students in Grades 8-10.

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