Math Problems: Solve For C, X, And Y

Dec 11, 2025 by Alex Johnson 37 views

Welcome, math enthusiasts! Today, we're diving into a couple of intriguing problems that will test your skills in algebra and unit conversion. We'll tackle a temperature conversion problem and then move on to a system of linear equations. Let's get started!

Problem 17: Converting Fahrenheit to Celsius

Our first challenge involves the ubiquitous formula for converting Fahrenheit to Celsius: $F = rac{5}{9}C + 32$ . You're given a Fahrenheit temperature, $F = 98.6$ , and your mission is to find the corresponding Celsius temperature, $C$ . This is a fantastic opportunity to practice rearranging algebraic formulas and substituting values. Remember, understanding how to manipulate equations is a fundamental skill in mathematics and science, allowing us to solve for unknown variables in countless real-world scenarios. Whether you're calculating body temperature, weather forecasts, or cooking temperatures, the ability to convert between Fahrenheit and Celsius is incredibly useful. We'll walk through this step-by-step, ensuring you grasp each part of the process.

To find $C$ when $F=98.6$ , we need to isolate $C$ in the given equation. First, let's substitute the value of $F$ into the formula:

$98.6 = rac{5}{9}C + 32$

Our goal is to get $C$ by itself on one side of the equation. The first step is to subtract 32 from both sides to isolate the term with $C$ :

$98.6 - 32 = rac{5}{9}C$

$66.6 = rac{5}{9}C$

Now, to get rid of the fraction $rac{5}{9}$ that's multiplying $C$ , we can multiply both sides of the equation by its reciprocal, which is $rac{9}{5}$ . This is a crucial step in algebraic manipulation, often referred to as "undoing" the operation. Since division by 5 and multiplication by 9 are currently applied to $C$ , multiplying by 9/5 effectively cancels them out.

$rac{9}{5} imes 66.6 = rac{9}{5} imes rac{5}{9}C$

On the right side, $rac{9}{5} imes rac{5}{9}$ cancels out to 1, leaving us with just $C$ . On the left side, we perform the multiplication:

$rac{9 imes 66.6}{5} = C$

$600.4 / 5 = C$

$119.2 = C$

So, when the temperature is 98.6 degrees Fahrenheit, it is equivalent to 119.2 degrees Celsius. This might seem a bit high for everyday temperatures, but it's a valid mathematical result. It's important to double-check your calculations, especially when dealing with decimals. A quick mental check: Celsius temperatures are generally lower than Fahrenheit temperatures, and 119.2 is significantly higher than 98.6, which at first glance might seem counter-intuitive. However, the formula dictates this conversion. Let's re-verify the calculation: $98.6 - 32 = 66.6$ . Then, $66.6 imes rac{9}{5}$ . $66.6 imes 9 = 599.4$ . And $599.4 / 5 = 119.88$ . Ah, a slight discrepancy! Let's re-do the multiplication more carefully. $98.6 = rac{5}{9}C + 32$ . $66.6 = rac{5}{9}C$ . $C = 66.6 imes rac{9}{5}$ . Let's convert 66.6 to a fraction: $66.6 = rac{666}{10} = rac{333}{5}$ . So, $C = rac{333}{5} imes rac{9}{5} = rac{2997}{25}$ . Now, let's perform the division: $2997 ig/ 25 = 119.88$ .

Therefore, $C = 119.88$ . This is a more accurate result. It's always good to be meticulous with arithmetic, especially with decimals. It highlights the importance of careful calculation in reaching the correct answer. This exercise in temperature conversion not only reinforces algebraic manipulation but also serves as a reminder of the precision required in mathematical operations.

Problem 18: Solving a System of Linear Equations

Moving on to our second problem, we encounter a classic system of linear equations:

$y + 2x = 4$
$y - 3x = -1$

Our goal is to find the value of $(x+y)$ . To do this, we first need to determine the individual values of $x$ and $y$ by solving the system. There are several methods to solve systems of equations, including substitution and elimination. We'll use the elimination method here, as it looks quite straightforward with the given equations. The elimination method is particularly useful when the coefficients of one of the variables are the same or can be easily made the same, allowing us to eliminate that variable by adding or subtracting the equations.

Notice that both equations have a ' $y$ ' term with a coefficient of +1. This makes elimination of ' $y$ ' very simple. We can subtract the second equation from the first equation to eliminate ' $y$ '. Let's write it out:

$(y + 2x) - (y - 3x) = 4 - (-1)$

Distribute the negative sign in the second parenthesis:

$y + 2x - y + 3x = 4 + 1$

Now, combine like terms. The ' $y$ ' terms cancel out ( $y - y = 0$ ), and the ' $x$ ' terms combine ( $2x + 3x = 5x$ ):

$5x = 5$

To find $x$ , we simply divide both sides by 5:

$x = rac{5}{5}$

$x = 1$

We've found the value of $x$ ! Now, we need to find the value of $y$ . We can substitute the value of $x=1$ into either of the original equations. Let's use the first equation: $y + 2x = 4$ .

Substitute $x=1$ into the equation:

$y + 2(1) = 4$

$y + 2 = 4$

Subtract 2 from both sides to solve for $y$ :

$y = 4 - 2$

$y = 2$

So, we have found that $x = 1$ and $y = 2$ . The problem asks for the value of $(x+y)$ . Now that we have $x$ and $y$ , we can easily calculate this:

$x + y = 1 + 2$

$x + y = 3$

Therefore, the value of $(x+y)$ is 3. This corresponds to option A. It's always a good practice to check our solution by substituting $x=1$ and $y=2$ back into both original equations to ensure they hold true.

For the first equation: $y + 2x = 2 + 2(1) = 2 + 2 = 4$ . This is correct. For the second equation: $y - 3x = 2 - 3(1) = 2 - 3 = -1$ . This is also correct.

Our solution is consistent with both equations, confirming that $x=1$ and $y=2$ are the correct values. This method of solving systems of equations is fundamental to many areas of mathematics and has wide-ranging applications in fields like economics, engineering, and computer science. The ability to find the intersection point of two lines (which is what solving a system of two linear equations represents graphically) is a powerful analytical tool.

Conclusion

We've successfully navigated two distinct mathematical challenges today! First, we converted a Fahrenheit temperature to Celsius using algebraic manipulation. Then, we tackled a system of linear equations to find the value of $x+y$ . These problems highlight the importance of precise calculation, algebraic fluency, and understanding different problem-solving strategies. Keep practicing these skills, and you'll find that mathematics becomes more accessible and enjoyable!

For further exploration into the world of mathematics and problem-solving, you might find these resources helpful:

Khan Academy: A fantastic free resource for learning and practicing math concepts from elementary school to college level. (www.khanacademy.org)
Wolfram Alpha: An incredible computational knowledge engine that can solve complex math problems and provide detailed explanations. (www.wolframalpha.com)