Rectangle Area, Equations & Age Puzzles Explained

Hey there, math enthusiasts! Today, we're diving into some exciting mathematical problems that involve rectangles, equations, and a bit of age-related fun. We'll be tackling questions about the area of a rectangle, solving for variables, and even figuring out how old someone will be in a few years. So, grab your thinking caps, and let's get started!

1. Decoding the Rectangle: Finding Length from Area and Breadth

Our first challenge involves a rectangle with an area of $(4x^2 - 6x) \text{ cm}^2$ and a breadth (or width) of $2x \text{ cm}$. The big question is: what's the length of this rectangle in terms of $x$? To solve this, we need to dust off our knowledge of rectangle area. Remember, the area of a rectangle is calculated by multiplying its length and breadth: $\text{Area} = \text{Length} \times \text{Breadth}$. In our case, we know the area and the breadth, and we're hunting for the length. So, we can rearrange the formula to solve for length: $\text{Length} = \frac{\text{Area}}{\text{Breadth}}$.

Now, let's plug in the values we know. The area is $(4x^2 - 6x) \text{ cm}^2$, and the breadth is $2x \text{ cm}$. This gives us: $ extLength} = \frac{4x^2 - 6x}{2x}$ To simplify this expression, we need to factor out the common factor in the numerator. Looking at $4x^2 - 6x$, we can see that both terms have a common factor of $2x$. Factoring out $2x$, we get $2x(2x - 3)$. So, our equation becomes $\text{Length = \frac2x(2x - 3)}{2x}$ Now, we can cancel out the $2x$ terms in the numerator and denominator, leaving us with $ ext{Length = 2x - 3$ Therefore, the length of the rectangle in terms of $x$ is $(2x - 3) \text{ cm}$. That wasn't so hard, was it? We used our understanding of area, rearranged the formula, factored out common terms, and simplified the expression to find the answer. This is a classic example of how algebraic manipulation can help us solve geometric problems.

Key Takeaways:

• The area of a rectangle is Length × Breadth.
• We can rearrange formulas to solve for unknown variables.
• Factoring out common terms simplifies algebraic expressions.

2. Cracking the Equation: Calculating 'y' When x = 4

Next up, we have an equation: $y = x^3 + 1$. Our mission, should we choose to accept it, is to calculate the value of $y$ when $x$ is equal to 4. This is a straightforward substitution problem, and it's a great way to practice our order of operations (PEMDAS/BODMAS). To solve for $y$, we simply replace $x$ with 4 in the equation: $y = (4)^3 + 1$ Now, we need to evaluate $4^3$. This means 4 multiplied by itself three times: $4 \times 4 \times 4$, which equals 64. So, our equation becomes: $y = 64 + 1$ Finally, we add 1 to 64, giving us: $y = 65$ So, when $x = 4$, the value of $y$ is 65. See how easy that was? We just substituted the value of $x$, followed the order of operations, and arrived at our answer. These types of problems are fundamental in algebra and help us understand how variables and equations work.

Key Takeaways:

• Substitution is a key technique for solving equations.
• Remember the order of operations (PEMDAS/BODMAS).
• Exponents indicate repeated multiplication.

3. The Age-Old Puzzle: Thandi and Sophie's Ages

Our final challenge involves a bit of a word problem. We're introduced to Thandi and Sophie, and we know that Thandi is six years older than Sophie. We also know that in three years, Thandi will be twice as old as Sophie. The question we need to answer is: how old are Thandi and Sophie right now? This type of problem requires us to translate the given information into algebraic equations and then solve them.

Let's start by assigning variables. Let Sophie's current age be $s$, and Thandi's current age be $t$. From the first piece of information, we know that Thandi is six years older than Sophie. This can be written as an equation: $t = s + 6$ Now, let's consider the second piece of information. In three years, Thandi will be $t + 3$ years old, and Sophie will be $s + 3$ years old. We're told that in three years, Thandi will be twice as old as Sophie. This gives us our second equation: $t + 3 = 2(s + 3)$ We now have a system of two equations with two variables:

$$t = s + 6$$

$$t + 3 = 2(s + 3)$$

To solve this system, we can use substitution. Since we already have an equation that expresses $t$ in terms of $s$ ($t = s + 6$), we can substitute this expression for $t$ in the second equation: $(s + 6) + 3 = 2(s + 3)$ Now, we have one equation with one variable, which we can solve. First, let's simplify both sides of the equation: $s + 9 = 2s + 6$ Next, we want to isolate the variable $s$. We can subtract $s$ from both sides: $9 = s + 6$ Then, subtract 6 from both sides: $3 = s$ So, Sophie is currently 3 years old. Now that we know Sophie's age, we can use the first equation ($t = s + 6$) to find Thandi's age: $t = 3 + 6$ $t = 9$ Therefore, Thandi is currently 9 years old. And there you have it! We've successfully solved the age puzzle by translating the word problem into algebraic equations and using substitution to find the solution. These types of problems are great for developing our problem-solving and critical-thinking skills.

Key Takeaways:

• Translate word problems into algebraic equations.
• Use variables to represent unknown quantities.
• Solve systems of equations using substitution or other methods.
• Always check your answers to make sure they make sense in the context of the problem.

Conclusion: Math is All Around Us!

We've tackled a variety of math problems today, from finding the length of a rectangle to solving equations and figuring out ages. These examples demonstrate how math concepts are applicable to real-world situations. By understanding these concepts and practicing our problem-solving skills, we can become more confident and capable mathematicians. Keep exploring, keep learning, and keep having fun with math!