Solving The Equation: 5/(s+3) = 10/(7s+2)

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Hey math whizzes and equation solvers! Today, we're diving deep into a cool problem that involves fractions and variables. We're going to solve the equation $\frac{5}{s+3}=\frac{10}{7 s+2}$. This might look a little intimidating at first glance with those fractions and the 's' hanging out in the denominators, but trust me, guys, once you break it down, it's totally manageable. We'll go step-by-step, and by the end of this, you'll feel like a total pro at tackling these kinds of algebraic puzzles. So, grab your thinking caps, maybe a coffee or your favorite study snack, and let's get this equation solved!

Understanding the Equation and Our Goal

Alright, let's look at our equation: $\frac{5}{s+3}=\frac{10}{7 s+2}$. Our main goal here, when we solve the equation $\frac{5}{s+3}=\frac{10}{7 s+2}$, is to find the value of 's' that makes both sides of the equation equal. Think of it like a balancing scale; we need to find that perfect weight 's' that keeps everything perfectly level. The reason we do this is to understand the relationship between the expressions on both sides. In mathematics, solving equations is fundamental. It helps us model real-world scenarios, from figuring out how much of an ingredient you need in a recipe to complex engineering problems. For this specific equation, we have two rational expressions (that's fancy talk for fractions with variables in them) set equal to each other. The key challenge here is dealing with the variables in the denominators. We can't just plug in any number for 's' because some numbers might make the denominator zero, which is a big no-no in math (division by zero is undefined!). So, we'll need to be mindful of any values of 's' that could cause this problem as we go.

The Cross-Multiplication Strategy

One of the most effective ways to solve the equation $\frac{5}{s+3}=\frac{10}{7 s+2}$ when you have two fractions set equal to each other is by using the cross-multiplication technique. It's a neat trick that helps us get rid of the denominators and turn our fractional equation into a simpler linear equation. How does it work, you ask? Well, imagine drawing lines connecting the numerator of one fraction to the denominator of the other, forming an 'X'. You then multiply the top number of the left fraction by the bottom number of the right fraction, and set that equal to the product of the bottom number of the left fraction and the top number of the right fraction. So, for our equation, this means we'll multiply 5 by $(7s+2)$ and set it equal to the product of $(s+3)$ and 10. This step is super crucial because it simplifies the problem dramatically. Instead of dealing with fractions, we'll soon be working with terms that are easier to combine and isolate the variable 's'. Remember, this method is valid as long as the denominators $(s+3)$ and $(7s+2)$ are not zero. We'll address those potential restrictions later, but for now, let's focus on the algebra. This cross-multiplication is a direct consequence of multiplying both sides of the equation by the common denominator, which effectively cancels out the original denominators. It's a shortcut that saves a lot of steps and is super handy for solving proportions like this one.

Performing the Multiplication

Now that we've decided to use cross-multiplication to solve the equation $\frac{5}{s+3}=\frac{10}{7 s+2}$, it's time to actually do the math. Following our cross-multiplication rule, we get:

$5 \times (7s + 2) = 10 \times (s + 3)$

This looks way less scary, right? We've successfully eliminated the fractions. The next step is to use the distributive property. This means we multiply the number outside the parentheses by each term inside the parentheses. On the left side, we multiply 5 by $7s$ and then by 2:

$5 \times 7s = 35s$ $5 \times 2 = 10$

So, the left side becomes $35s + 10$.

Now, let's tackle the right side. We multiply 10 by $s$ and then by 3:

$10 \times s = 10s$ $10 \times 3 = 30$

So, the right side becomes $10s + 30$.

Putting it all together, our equation now looks like this:

$35s + 10 = 10s + 30$

See? We've transformed the original fractional equation into a nice, clean linear equation. This is where the real fun of isolating the variable begins. Each step we take here is designed to get 's' all by itself on one side of the equation. This process of simplifying and distributing is a core skill in algebra, and practicing it will make you incredibly efficient at solving problems. Keep going, guys, you're doing great!

Isolating the Variable 's'

We're at a fantastic stage now, having simplified our equation to $35s + 10 = 10s + 30$. Our next mission, should we choose to accept it (and we totally should!), is to solve the equation $\frac{5}{s+3}=\frac{10}{7 s+2}$ by isolating the variable 's'. This means we want to get all the terms containing 's' on one side of the equation and all the constant terms (the numbers without 's') on the other side. It's like tidying up your room – get all the 's' stuff in one corner and all the number stuff in another. Let's start by moving the 's' terms. I usually prefer to move the smaller 's' term to avoid dealing with negative coefficients, but either way works. So, let's subtract $10s$ from both sides of the equation. This is a key algebraic move; whatever you do to one side, you must do to the other to maintain the equality.

$35s - 10s + 10 = 10s - 10s + 30$

This simplifies to:

$25s + 10 = 30$

Awesome! Now, all our 's' terms are on the left side. The next step is to move the constant term, which is '+10' on the left side, to the right side. To do this, we'll subtract 10 from both sides:

$25s + 10 - 10 = 30 - 10$

This leaves us with:

$25s = 20$

We're so close! We've successfully grouped all the 's' terms together and all the constant terms together. The equation is now in its simplest form, $25s = 20$. This is a huge accomplishment and means we're just one tiny step away from finding the value of 's'. Keep that focus, and we'll nail this!

Finding the Final Value of 's'

We've reached the final stretch in our quest to solve the equation $\frac{5}{s+3}=\frac{10}{7 s+2}$! Our equation is currently $25s = 20$. To find the value of 's', we need to get 's' completely by itself. Right now, 's' is being multiplied by 25. The opposite operation of multiplication is division. So, to isolate 's', we need to divide both sides of the equation by 25. Remember, we have to do the same thing to both sides to keep the equation balanced.

$\frac{25s}{25} = \frac{20}{25}$

On the left side, the 25s cancel out, leaving us with just 's'. On the right side, we have the fraction $\frac{20}{25}$. This fraction can be simplified. Both 20 and 25 are divisible by 5. So, we divide both the numerator and the denominator by 5:

$20 \div 5 = 4$ $25 \div 5 = 5$

So, the simplified fraction is $\frac{4}{5}$. Therefore, our solution is:

$s = \frac{4}{5}$

Bingo! We've found the value of 's' that satisfies the original equation. It's $\frac{4}{5}$. This means that if you were to substitute $\frac{4}{5}$ back into the original equation for 's', both sides would be equal. Pretty neat, huh? This step of simplifying fractions is also super important. Always try to reduce your fractions to their simplest form, it makes the answer look cleaner and is often required in math problems.

Checking Our Solution

It's always a good idea, especially when you're learning or want to be absolutely sure, to check our solution for the equation $\frac{5}{s+3}=\frac{10}{7 s+2}$. We found that $s = \frac{4}{5}$. Let's plug this value back into the original equation and see if the left side equals the right side. This verification step is crucial for confirming our calculations and building confidence in our algebraic skills.

Left Side: $\frac{5}{s+3} = \frac{5}{\frac{4}{5}+3}$

To add $\frac{4}{5}$ and 3, we need a common denominator. Since 3 can be written as $\frac{15}{5}$ (because $3 \times 5 = 15$), we have: $\frac{4}{5} + \frac{15}{5} = \frac{4+15}{5} = \frac{19}{5}$

So, the left side becomes: $\frac{5}{\frac{19}{5}}$

Dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of $\frac{19}{5}$ is $\frac{5}{19}$. $5 \times \frac{5}{19} = \frac{25}{19}$

Right Side: $\frac{10}{7 s+2} = \frac{10}{7 \times \frac{4}{5}+2}$

First, calculate $7 \times \frac{4}{5}$: $7 \times \frac{4}{5} = \frac{28}{5}$

Now add 2 to $\frac{28}{5}$. We write 2 as $\frac{10}{5}$ (because $2 \times 5 = 10$). $\frac{28}{5} + \frac{10}{5} = \frac{28+10}{5} = \frac{38}{5}$

So, the right side becomes: $\frac{10}{\frac{38}{5}}$

Again, we divide by the fraction by multiplying by its reciprocal. The reciprocal of $\frac{38}{5}$ is $\frac{5}{38}$. $10 \times \frac{5}{38} = \frac{50}{38}$

Now, we simplify $\frac{50}{38}$. Both numbers are divisible by 2. $50 \div 2 = 25$ $38 \div 2 = 19$

So, the right side simplifies to $\frac{25}{19}$.

Comparison: Left Side = $\frac{25}{19}$ Right Side = $\frac{25}{19}$

They match! This confirms that our solution $s = \frac{4}{5}$ is correct. It's so satisfying when the numbers work out perfectly, right? Always take the time to check your answers; it's a small step that makes a big difference in accuracy.

Important Considerations: Domain Restrictions

Before we wrap up, it's super important, guys, to talk about domain restrictions when we solve the equation $\frac{5}{s+3}=\frac{10}{7 s+2}$. Remember how we said division by zero is undefined? Well, in our original equation, the denominators are $(s+3)$ and $(7s+2)$. If either of these becomes zero for a particular value of 's', then that value of 's' is not allowed in our solution set. We call these