Solving Linear Equations: How Many Solutions Exist?

by ADMIN 52 views

Hey guys! Ever wondered how to figure out if a math equation has one solution, infinite solutions, or no solution at all? Well, let's dive into a super common type of problem: determining the number of solutions for a linear equation. We're going to break down a specific equation, 13bβˆ’bβˆ’11=19+14b13b - b - 11 = 19 + 14b, and walk through the steps to find out exactly how many solutions it has. So, grab your thinking caps, and let's get started!

Understanding Linear Equations

Before we jump into solving, let's quickly recap what a linear equation is. A linear equation is basically an equation where the highest power of the variable (in our case, 'b') is 1. These equations, when graphed, form a straight lineβ€”hence the name "linear." They're the bread and butter of algebra, and understanding them is crucial for more advanced math. The general form of a linear equation is something like ax+b=cax + b = c, where 'x' is the variable and 'a', 'b', and 'c' are constants. Our goal is to find the value(s) of the variable that make the equation true.

Types of Solutions

Linear equations can have three possible types of solutions:

  1. One Unique Solution: This is the most common scenario. You simplify the equation and find a single value for the variable that satisfies the equation. For example, x=5x = 5 is a unique solution.
  2. Infinite Solutions: This happens when the equation simplifies to a statement that is always true, regardless of the value of the variable. For example, 0=00 = 0 or 5=55 = 5. This means any value you plug in for the variable will make the equation true.
  3. No Solution: This occurs when the equation simplifies to a statement that is always false. For example, 0=10 = 1 or 2=32 = 3. No matter what value you substitute for the variable, the equation will never be true.

Solving the Equation 13bβˆ’bβˆ’11=19+14b13b - b - 11 = 19 + 14b

Okay, let's tackle our equation: 13bβˆ’bβˆ’11=19+14b13b - b - 11 = 19 + 14b. Our mission is to simplify it and see which of the three solution scenarios we end up with.

Step 1: Simplify Both Sides

First, we'll simplify each side of the equation by combining like terms. On the left side, we have 13bβˆ’b13b - b, which simplifies to 12b12b. So the left side becomes 12bβˆ’1112b - 11. The right side, 19+14b19 + 14b, is already simplified.

So now our equation looks like this:

12bβˆ’11=19+14b12b - 11 = 19 + 14b

Step 2: Get the Variable Terms on One Side

Next, we want to get all the terms with 'b' on one side of the equation. Let's subtract 14b14b from both sides to move the 'b' terms to the left side:

12bβˆ’11βˆ’14b=19+14bβˆ’14b12b - 11 - 14b = 19 + 14b - 14b

This simplifies to:

βˆ’2bβˆ’11=19-2b - 11 = 19

Step 3: Isolate the Variable Term

Now, we want to isolate the term with 'b'. To do this, we'll add 11 to both sides of the equation:

βˆ’2bβˆ’11+11=19+11-2b - 11 + 11 = 19 + 11

This simplifies to:

βˆ’2b=30-2b = 30

Step 4: Solve for 'b'

Finally, to solve for 'b', we'll divide both sides of the equation by -2:

βˆ’2bβˆ’2=30βˆ’2\frac{-2b}{-2} = \frac{30}{-2}

This gives us:

b=βˆ’15b = -15

Determining the Number of Solutions

Alright, we've crunched the numbers and found that b=βˆ’15b = -15. This means that there is one unique value for 'b' that makes the original equation true. Therefore, the equation 13bβˆ’bβˆ’11=19+14b13b - b - 11 = 19 + 14b has one unique solution.

Checking Our Solution

To be absolutely sure, we can plug our solution, b=βˆ’15b = -15, back into the original equation and see if it holds true:

13(βˆ’15)βˆ’(βˆ’15)βˆ’11=19+14(βˆ’15)13(-15) - (-15) - 11 = 19 + 14(-15)

βˆ’195+15βˆ’11=19βˆ’210-195 + 15 - 11 = 19 - 210

βˆ’191=βˆ’191-191 = -191

Since both sides of the equation are equal, our solution is correct!

Scenarios: One Solution, Infinite Solutions, or No Solution

Let's briefly discuss how different outcomes would indicate different numbers of solutions. Remember, we're looking for patterns that tell us whether we have one, infinite, or no solutions.

One Unique Solution

As we saw in our example, when you solve the equation and find a specific value for the variable (e.g., b=βˆ’15b = -15), you have one unique solution. This means there is only one value that, when plugged into the original equation, will make it true.

Infinite Solutions

Imagine we ended up with an equation that looked like this after simplification:

0=00 = 0

Or something like:

2x+4=2x+42x + 4 = 2x + 4

In these cases, no matter what value you substitute for 'x', the equation will always be true. This is because both sides of the equation are identical. When this happens, we say the equation has infinite solutions. Any real number will satisfy the equation.

No Solution

Now, suppose we simplified an equation and ended up with something like this:

0=50 = 5

Or:

x=x+1x = x + 1

These statements are clearly false. No matter what value you plug in for 'x', the equation will never be true. In this case, we say the equation has no solution. There is no value that can satisfy the equation.

Tips for Solving Linear Equations

Solving linear equations can be straightforward if you follow a few key steps. Here are some tips to help you along the way:

  • Simplify First: Always start by simplifying both sides of the equation. Combine like terms and distribute any terms as needed. This will make the equation easier to work with.
  • Isolate the Variable: Get all the terms with the variable on one side of the equation and all the constants on the other side. Use addition and subtraction to move terms around.
  • Solve for the Variable: Once the variable is isolated, divide or multiply to solve for the variable. Remember to perform the same operation on both sides of the equation to maintain balance.
  • Check Your Solution: After you find a solution, plug it back into the original equation to make sure it is correct. This is a great way to catch any mistakes.
  • Stay Organized: Keep your work neat and organized. Write down each step clearly so you can easily follow your reasoning. This will help you avoid mistakes and make it easier to check your work.

Conclusion

So, to recap, the equation 13bβˆ’bβˆ’11=19+14b13b - b - 11 = 19 + 14b has one unique solution, which we found to be b=βˆ’15b = -15. By simplifying the equation and isolating the variable, we were able to determine the exact value that satisfies the equation. Remember to always simplify, isolate, and solve, and don't forget to check your answer! Understanding how to determine the number of solutions for a linear equation is a fundamental skill in algebra, and with practice, you'll become a pro at solving these types of problems. Keep practicing, and you'll master these skills in no time! You got this!