Solving Linear Equations By Substitution: Step-by-Step

Hey guys! Today, we're diving deep into the world of linear equations and tackling them using the substitution method. If you've ever felt lost trying to solve these, don't worry, you're in the right place. We'll break down the process step-by-step and make sure you've got a solid understanding of how to conquer these problems. We will solve the next two questions in detail:

i. $y+\frac{1}{y}=2 \frac{1}{2}$ ii. $\frac{x 10}{x-5}=\frac{7 x}{x-5}$

So, grab your pencils, notebooks, and let's get started!

What is the Substitution Method?

Before we jump into specific examples, let's quickly recap what the substitution method actually is. In essence, the substitution method is a powerful technique for solving systems of equations. Instead of trying to eliminate variables (like in the elimination method), we focus on isolating one variable in one equation and then substituting that expression into the other equation. This effectively reduces the problem to a single equation with a single variable, which is much easier to solve. Think of it like this: you're finding an equivalent value for a variable and then plugging it into another equation to simplify things.

The beauty of the substitution method lies in its versatility. It's particularly useful when one of the equations has a variable that's already isolated or can be easily isolated. But don't let that fool you – it's a valuable tool in your problem-solving arsenal, no matter the complexity of the equations. It is crucial to understand that sometimes you might need to rearrange the equations to make the substitution method work best. This might involve adding, subtracting, multiplying, or dividing terms to isolate a variable. However, this initial effort can save you a lot of trouble in the long run. Remember, the goal is to make the problem as simple as possible so you can solve it efficiently and accurately. Also, checking your solutions is always a good practice. Once you've found a solution, plug it back into the original equations to verify that it satisfies both. This helps prevent errors and gives you confidence in your answer.

This method is extremely helpful when we are dealing with complex equations where direct solutions might not be immediately apparent. By breaking down the problem into smaller, manageable parts, substitution makes it easier to tackle even the most daunting systems of equations. You will find this method invaluable not just in your math classes but also in various real-world applications where you need to solve multiple interrelated variables. So, pay close attention, practice diligently, and soon you'll be a pro at solving equations using substitution!

Example 1: Solving $y+\frac{1}{y}=2 \frac{1}{2}$

Let's start with our first equation: $y+\frac{1}{y}=2 \frac{1}{2}$. This might look a bit intimidating at first, but trust me, we can handle it. The key here is to get rid of that fraction and transform this into a more familiar quadratic equation. So, how do we do that? We'll multiply both sides of the equation by y. This will eliminate the fraction and make the equation much easier to work with. This is a common trick in algebra – if you see a fraction causing trouble, try multiplying through by the denominator.

So, multiplying both sides by y, we get:

$y(y + \frac{1}{y}) = y(2 \frac{1}{2})$

This simplifies to:

$y^2 + 1 = \frac{5}{2}y$

Now, let's get rid of that fraction on the right side. We can do this by multiplying both sides of the equation by 2:

$2(y^2 + 1) = 2(\frac{5}{2}y)$

Which gives us:

$2y^2 + 2 = 5y$

Alright, now we've got a quadratic equation! To solve it, we need to rearrange it into the standard form, which is $ax^2 + bx + c = 0$. In our case, that means subtracting 5y from both sides:

$2y^2 - 5y + 2 = 0$

Now we have a classic quadratic equation. There are a couple of ways we can solve this: factoring or using the quadratic formula. Let's try factoring first, as it can be quicker if the equation factors nicely. Factoring involves finding two binomials that multiply together to give us our quadratic equation. We need to find two numbers that multiply to give 2 * 2 = 4 and add up to -5. Those numbers are -4 and -1. So, we can rewrite the middle term as -4y - y:

$2y^2 - 4y - y + 2 = 0$

Now, we can factor by grouping. We group the first two terms and the last two terms:

$(2y^2 - 4y) + (-y + 2) = 0$

Factor out the greatest common factor from each group:

$2y(y - 2) - 1(y - 2) = 0$

Notice that we now have a common factor of (y - 2). We can factor that out:

$(2y - 1)(y - 2) = 0$

Now, we can use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. So, we set each factor equal to zero:

$2y - 1 = 0$ or $y - 2 = 0$

Solving for y in each equation, we get:

$y = \frac{1}{2}$ or $y = 2$

And there you have it! We've found our solutions for y: $\frac{1}{2}$ and 2. Remember, it's always a good idea to plug these values back into the original equation to check that they work. In this case, both solutions satisfy the original equation, so we're confident in our answer.

Example 2: Solving $\frac{x 10}{x-5}=\frac{7 x}{x-5}$

Okay, let's move on to our second equation: $\frac{x + 10}{x-5}=\frac{7x}{x-5}$. This one looks like it might involve some tricky fractions, but don't worry, we'll tackle it together. The first thing we should notice is that both fractions have the same denominator, (x - 5). This is a big clue that we can simplify things quite a bit. Whenever you see the same denominator in an equation like this, your first thought should be to get rid of it. How do we do that? Simple – we multiply both sides of the equation by that denominator.

So, let's multiply both sides by (x - 5):

$(x - 5)(\frac{x + 10}{x-5}) = (x - 5)(\frac{7x}{x-5})$

Notice what happens on both sides: the (x - 5) terms cancel out! This is exactly what we wanted. We're left with:

$x + 10 = 7x$

Wow, that's much simpler, isn't it? Now we have a basic linear equation to solve. Our goal is to isolate x on one side of the equation. Let's start by subtracting x from both sides:

$x + 10 - x = 7x - x$

This simplifies to:

$10 = 6x$

Now, to get x by itself, we divide both sides by 6:

$\frac{10}{6} = x$

We can simplify the fraction $\frac{10}{6}$ by dividing both the numerator and the denominator by their greatest common factor, which is 2:

$x = \frac{5}{3}$

So, we've found our solution: $x = \frac{5}{3}$. But before we declare victory, there's one crucial thing we need to check. Remember that original equation with the fractions? We had a denominator of (x - 5). We need to make sure that our solution doesn't make that denominator equal to zero, because division by zero is undefined. Let's plug our solution, $\frac{5}{3}$, back into (x - 5):

$\frac{5}{3} - 5 = \frac{5}{3} - \frac{15}{3} = -\frac{10}{3}$

Since this is not zero, our solution is valid. If it had been zero, we would have had to discard the solution and potentially look for other solutions or conclude that there is no solution. This step is super important – it's a common mistake to forget to check for these extraneous solutions, especially when dealing with rational equations (equations with fractions).

So, to recap, we started with a fractional equation, multiplied both sides by the common denominator to eliminate the fractions, solved the resulting linear equation, and then checked our solution to make sure it was valid. By following these steps carefully, you can confidently tackle similar problems.

Tips and Tricks for Mastering the Substitution Method

Alright, now that we've worked through a couple of examples, let's talk about some tips and tricks that can help you master the substitution method. These are the little things that can make a big difference in your problem-solving efficiency and accuracy. First off, always look for the easiest variable to isolate. When you're deciding which equation to start with and which variable to isolate, choose the one that requires the fewest steps. This often means looking for a variable that already has a coefficient of 1 or -1. Isolating such a variable is usually straightforward and avoids unnecessary fractions. For instance, if you have the equations x + 2y = 5 and 3x - y = 1, isolating y in the second equation is a good starting point because it only involves adding y to both sides and subtracting 1 from both sides.

Another crucial tip is to be careful with signs, guys! Sign errors are super common in algebra, especially when dealing with negative numbers. When you're substituting an expression, make sure you distribute any negative signs correctly. For example, if you're substituting (2 - x) into an equation, remember that -(2 - x) is -2 + x, not -2 - x. Taking your time and double-checking your work can save you from making these kinds of mistakes. Also, don't be afraid to rewrite equations to make them clearer. Sometimes, just rearranging the terms can help you visualize the problem better and reduce the chance of errors. For example, if you have an equation like 5 - x = 2y, you might want to rewrite it as -x = 2y - 5 and then as x = -2y + 5. This can make the isolation process more intuitive.

Remember to check your solutions in both original equations. We talked about this earlier, but it's worth repeating: always, always, always check your solutions by plugging them back into the original equations. This ensures that your solutions are correct and that you haven't made any mistakes along the way. It also helps you catch extraneous solutions, as we saw in our second example. And finally, practice makes perfect! The more you practice using the substitution method, the more comfortable you'll become with it. Work through lots of different examples, and don't be afraid to tackle challenging problems. With enough practice, you'll be able to solve systems of equations with confidence.

Common Mistakes to Avoid

Now, let's talk about some common mistakes people make when using the substitution method. Knowing these pitfalls can help you steer clear of them and improve your accuracy. One of the biggest mistakes is incorrect distribution of signs, we already mentioned that. This often happens when substituting a negative expression. Another frequent error is forgetting to substitute back into both original equations. Remember, a solution to a system of equations must satisfy all equations in the system. If you only check your solution in one equation, you might miss an error in the other equation.

Another mistake is choosing the harder variable to isolate. We talked about looking for the easiest variable to isolate, but sometimes people overlook this and end up doing more work than necessary. If you find yourself dealing with complicated fractions or multiple steps, take a step back and see if there's a simpler way to approach the problem. Also, be careful when dealing with fractions. Fractions can be tricky, and it's easy to make mistakes if you're not careful. If you're uncomfortable working with fractions, consider clearing them by multiplying both sides of the equation by the least common denominator. This can make the equation much easier to work with.

Also, sometimes students make mistakes while simplifying the equation after substitution. After you substitute, you'll have a new equation with one variable. Make sure you simplify this equation correctly by combining like terms and performing the necessary operations. Double-check your work to avoid errors. And lastly, don't forget to check for extraneous solutions, especially when dealing with rational equations or equations with radicals. As we saw in our example, extraneous solutions can arise when we perform operations that are not reversible, such as squaring both sides of an equation. Checking your solutions is the only way to catch these extraneous solutions.

By being aware of these common mistakes and taking steps to avoid them, you can significantly improve your accuracy and confidence when using the substitution method. Remember, math is all about practice and attention to detail. The more you practice, the better you'll become at spotting and avoiding these errors.

Conclusion

So, there you have it, guys! We've covered the substitution method for solving linear equations in detail. We've walked through examples, discussed tips and tricks, and highlighted common mistakes to avoid. With this knowledge, you're well-equipped to tackle a wide range of problems. Remember, the key to mastering any mathematical technique is practice. Work through lots of examples, and don't be afraid to ask for help if you get stuck. Keep practicing, and soon you'll be solving linear equations like a pro! Happy solving!