If the length of a rectangle is increased by 5 metres and the breadth decreased by 3 metres, the area would decrease by 5 square metres. If the length is increased by 3 metres and breadth increased by 2 metres, the area would increase by 50 square metres. What are the length and breadth?

Let the length of the rectangle be 'x' metres and the breadth of the rectangle be 'y' metres. Area = xy square metre.

To solve the problem, we need to set up a system of equations based on the information given about the changes in dimensions and how they impact the area of the rectangle.

Let's define:

Original length of the rectangle as [tex]x[/tex] meters.
Original breadth of the rectangle as [tex]y[/tex] meters.
The original area is then [tex]xy[/tex] square meters.

Step 1: Formulating the first condition.

According to the first condition:

Length increases by 5 meters, so new length = [tex]x + 5[/tex] meters.
Breadth decreases by 3 meters, so new breadth = [tex]y - 3[/tex] meters.
New area = [tex](x + 5)(y - 3)[/tex].
The area decreases by 5 square meters, which gives:

[tex](x + 5)(y - 3) = xy - 5[/tex]

Simplifying this gives:

[tex]xy - 3x + 5y - 15 = xy - 5[/tex]

Now, cancel out [tex]xy[/tex] from both sides and simplify:

[tex]-3x + 5y - 15 = -5[/tex]

Adding 15 to both sides, we have:

[tex]-3x + 5y = 10 \quad \text{(Equation 1)}[/tex]

Step 2: Formulating the second condition.

From the second condition:

Length increases by 3 meters, so the length = [tex]x + 3[/tex] meters.
Breadth increases by 2 meters, so the breadth = [tex]y + 2[/tex] meters.
The area increases by 50 square meters, which gives:

[tex](x + 3)(y + 2) = xy + 50[/tex]

Simplifying:

[tex]xy + 2x + 3y + 6 = xy + 50[/tex]

Cancel [tex]xy[/tex] and simplify:

[tex]2x + 3y + 6 = 50[/tex]

Subtract 6 from both sides:

[tex]2x + 3y = 44 \quad \text{(Equation 2)}[/tex]

Step 3: Solving the simultaneous equations.

We have now two equations:

Equation 1: [tex]-3x + 5y = 10[/tex]
Equation 2: [tex]2x + 3y = 44[/tex]

We can solve these equations using any method, such as substitution or elimination. Here, let's use the elimination method:

To eliminate [tex]x[/tex], multiply Equation 1 by 2 and Equation 2 by 3:

[tex]\begin{align*}
-6x + 10y &= 20 \
6x + 9y &= 132 \
\end{align*}[/tex]

Add these equations:

[tex]19y = 152[/tex]

Solving for [tex]y[/tex]:

[tex]y = \frac{152}{19} = 8[/tex]

Substitute [tex]y = 8[/tex] back into Equation 2:

[tex]2x + 3(8) = 44[/tex]

[tex]2x + 24 = 44[/tex]

Subtract 24 from both sides:

[tex]2x = 20[/tex]

Divide by 2:

[tex]x = 10[/tex]

The length of the rectangle is [tex]10[/tex] meters, and the breadth is [tex]8[/tex] meters.