Solve System Of Equations 3x+y=9 And 3x-5y=15 A Step-by-Step Guide
Hey there, math enthusiasts! Today, we are diving into a classic algebra problem: solving a system of linear equations. Specifically, we are tackling the following system:
We've got a multiple-choice question here, and our mission is to find the correct solution for x and y. Don't worry; we'll break it down step by step, so it's super clear and easy to follow. Let's get started!
Understanding Systems of Equations
Before we jump into solving, let's quickly recap what a system of equations is. A system of equations is simply a set of two or more equations containing the same variables. The solution to a system of equations is the set of values for the variables that make all the equations true simultaneously. In our case, we have two equations and two variables (x and y), so we are looking for the unique pair of x and y values that satisfy both equations.
There are several methods to solve systems of equations, including substitution, elimination, and graphing. For this particular problem, the elimination method seems like a great fit because we notice that the coefficients of the x terms in both equations are the same (both are 3). This means we can easily eliminate the x variable by subtracting one equation from the other. How cool is that?
The Elimination Method: A Step-by-Step Guide
The elimination method involves manipulating the equations in the system so that when you add or subtract them, one of the variables is eliminated. This leaves you with a single equation in one variable, which you can then easily solve. Let's apply this to our system:
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Write down the equations:
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Identify the variable to eliminate: As we discussed, the x terms have the same coefficient (3), so we'll eliminate x. This is where the magic begins, guys!
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Subtract the second equation from the first equation:
Be super careful with the signs here! Distribute the negative sign properly:
Notice how the 3x and -3x cancel each other out, eliminating x:
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Solve for y: Now we have a simple equation with just y. Divide both sides by 6 to isolate y:
Awesome! We've found the value of y. y equals -1. This is a major step forward. Feels good, doesn't it?
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Substitute the value of y back into one of the original equations to solve for x: We can use either equation, but let's use the first one, $3x + y = 9$, because it looks a bit simpler:
Add 1 to both sides:
Divide both sides by 3:
Fantastic! We've also found the value of x. x equals 10/3. We are on fire, folks!
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Write the solution as an ordered pair: The solution to the system of equations is the ordered pair (x, y), which in our case is (10/3, -1).
Checking Our Solution
It's always a good idea to check your solution to make sure you didn't make any mistakes. To do this, substitute the values of x and y back into both original equations and see if they hold true.
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Equation 1: $3x + y = 9$
9 = 9$ **(True!)**
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Equation 2: $3x - 5y = 15$
15 = 15$ **(True!)**
Both equations are true, so our solution is correct. Woohoo! We nailed it!
Comparing Our Solution to the Options
Now let's look at the multiple-choice options:
A. $x=\frac{10}{3}, y=-1$ B. $x=\frac{8}{3}, y=1$ C. $x=-1, y=\frac{10}{3}$ D. $x=1, y=\frac{8}{3}$
Our solution, (10/3, -1), matches option A. So, the correct answer is A.
Why the Other Options Are Incorrect
It's helpful to understand why the other options are wrong. This not only confirms our answer but also deepens our understanding of the problem. Let's briefly examine why options B, C, and D are incorrect.
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Option B: $x=\frac{8}{3}, y=1$
If we substitute these values into the first equation, $3x + y = 9$, we get:
9 = 9$ **(True for the first equation)** But let's check the second equation, $3x - 5y = 15$: $3(\frac{8}{3}) - 5(1) = 15
3 = 15$ **(False!)** Since option B doesn't satisfy both equations, it's incorrect.
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Option C: $x=-1, y=\frac{10}{3}$
Substituting these values into the first equation, $3x + y = 9$, gives us:
\frac{1}{3} = 9$ **(False!)** Option C fails the first equation itself, so it's incorrect.
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Option D: $x=1, y=\frac{8}{3}$
Plugging these values into the first equation, $3x + y = 9$, we get:
\frac{17}{3} = 9$ **(False!)** Option D also fails the first equation and is therefore incorrect.
By checking these options, we see definitively that only option A satisfies both equations in the system.
Key Takeaways for Solving Systems of Equations
Before we wrap up, let's highlight some key takeaways from this problem-solving journey. These tips will help you tackle similar problems with confidence:
- Understand the concept: Make sure you grasp what a system of equations represents β a set of equations with common variables, and the solution is the set of values that make all equations true.
- Choose the right method: Decide which method (substitution, elimination, or graphing) is most efficient for the given system. In our case, elimination was ideal due to the matching coefficients of x.
- Be careful with signs: When using elimination, pay close attention to signs, especially when subtracting equations. A small sign error can lead to a wrong answer.
- Substitute correctly: After solving for one variable, substitute its value back into one of the original equations to find the other variable. Double-check your substitution to avoid errors.
- Check your solution: Always verify your solution by substituting the values of the variables back into both original equations. This ensures accuracy and builds confidence.
- Understand why incorrect options are wrong: Analyzing the incorrect options can reinforce your understanding of the problem and the solution process.
Practice Makes Perfect
Solving systems of equations is a fundamental skill in algebra, and like any skill, it improves with practice. Try solving more problems using different methods, and you'll become a pro in no time. Guys, you've got this!
Remember, math isn't about memorizing formulas; it's about understanding concepts and applying them logically. Keep practicing, keep exploring, and keep enjoying the beauty of mathematics!
So, to recap, the solution to the system of equations $3x + y = 9$ and $3x - 5y = 15$ is indeed A. $x=\frac{10}{3}, y=-1$. Great job, everyone! Keep up the excellent work, and I'll see you in the next math adventure! Now you're equipped to tackle similar problems and impress your friends with your mad algebra skills!
Understanding the Problem and the Question
Okay, let's break down this math problem step by step, making sure we understand exactly what's being asked. Our core question here is: What is the solution to the system of equations
and
3x - 5y = 15$? This is a classic algebra question that involves finding the values of *x* and *y* that satisfy both equations simultaneously. Think of it as a puzzle where we need to find the perfect *x* and *y* combination that fits both pieces of information. There are a few different methods we can use to solve this, but we'll walk through one of the most common and efficient methods β the elimination method. This will help you tackle similar problems with confidence. Remember, the goal is not just to get the answer but to understand the process, so you can apply these skills to other mathematical challenges. ### Breaking Down the Equations Before we dive into solving, let's take a closer look at what these equations actually represent. Each equation is a linear equation, which means that if we were to graph them, they would form straight lines. The solution to the system of equations is the point where these two lines intersect on a graph. At that point of intersection, the *x* and *y* values will satisfy both equations. So, we're essentially looking for the coordinates of this intersection point without actually having to graph the lines. Pretty neat, huh? Linear equations are powerful tools in mathematics and have many real-world applications. They are used to model relationships between two variables that change at a constant rate. For example, you might use a linear equation to represent the cost of renting a car based on the number of miles you drive, or the distance a train travels over time at a constant speed. Understanding how to solve systems of linear equations is a fundamental skill that opens doors to more advanced mathematical concepts and problem-solving techniques. It's like learning the basic chords on a guitar before you can play a whole song! In our specific case, the equations $3x + y = 9$ and $3x - 5y = 15$ give us two different relationships between *x* and *y*. The first equation tells us that three times the value of *x* plus the value of *y* equals 9. The second equation tells us that three times the value of *x* minus five times the value of *y* equals 15. Our job is to find the single pair of *x* and *y* values that make both of these statements true at the same time. This is where the magic of algebra comes in! ### Multiple Choice Options: A Quick Overview We've been given four multiple-choice options for the solution: A. $x=\frac{10}{3}, y=-1
B. $x=\frac{8}{3}, y=1$
C. $x=-1, y=\frac{10}{3}$
D. $x=1, y=\frac{8}{3}$
Each option provides a potential pair of x and y values. Our goal is to determine which of these pairs, if any, satisfies both equations in our system. One way to approach this is to simply plug in the x and y values from each option into the equations and see if they hold true. However, a more efficient method is to solve the system of equations directly and then compare our solution to the options. This ensures that we not only find the correct answer but also understand the process thoroughly. Think of it like this: we could try each key on a keychain until we find the one that opens the lock, or we could learn how the lock works and figure out exactly which key is needed. The latter approach is much more powerful in the long run!
The Elimination Method: A Detailed Explanation
Alright, let's dive into the elimination method, which is a super effective way to solve systems of equations like this one. The basic idea behind the elimination method is to manipulate the equations in such a way that when we add or subtract them, one of the variables cancels out. This leaves us with a single equation in one variable, which we can then easily solve. Once we have the value of one variable, we can substitute it back into one of the original equations to find the value of the other variable. It's like a strategic game of subtraction that leads us to the solution!
Step 1: Setting Up for Elimination
In our system of equations,
We notice something very convenient: the coefficients of the x terms are the same in both equations (they're both 3). This makes the elimination method particularly straightforward. If the coefficients were different, we might need to multiply one or both equations by a constant to make the coefficients match. But in this case, we're already set to go! This is like finding a puzzle piece that fits perfectly without needing to be rotated or flipped. Sometimes math problems give you a little break, and this is one of those times!
Step 2: Eliminating x
Since the coefficients of x are the same, we can eliminate x by subtracting the second equation from the first equation. This is a crucial step, so let's do it carefully. We'll subtract the entire left side of the second equation from the left side of the first equation, and we'll do the same for the right sides. Remember, it's all about keeping the balance! When subtracting equations, itβs important to distribute the negative sign correctly. This is a common area where mistakes can happen, so let's be extra vigilant. We're essentially performing the following operation:
Now, let's distribute the negative sign and simplify:
Notice how the 3x and -3x terms cancel each other out, as planned! This is the magic of the elimination method in action. We've successfully eliminated x from the equation. What remains is:
We've reduced our system of two equations with two variables into a single equation with just one variable. This is a significant step forward!
Step 3: Solving for y
Now that we have the equation $6y = -6$, solving for y is a breeze. We simply divide both sides of the equation by 6:
Fantastic! We've found the value of y. It's -1. This is a major milestone in our problem-solving journey. We're halfway to the solution! With y in hand, we can now move on to finding x.
Step 4: Substituting to Find x
To find the value of x, we'll substitute the value we just found for y (which is -1) back into one of the original equations. We can choose either equation; it doesn't matter which one we pick. However, it's often a good strategy to choose the simpler equation to minimize the chances of making a mistake. In this case, the first equation, $3x + y = 9$, looks a bit easier to work with, so let's go with that one.
Substituting y = -1 into the equation, we get:
Now, we just need to solve for x. We'll add 1 to both sides of the equation:
And then divide both sides by 3:
Excellent! We've found the value of x. It's $\frac{10}{3}$. We now have both x and y values that satisfy the system of equations. We're on the verge of solving this puzzle!
Step 5: The Solution and Verification
We've found that x = $\frac{10}{3}$ and y = -1. So, the solution to the system of equations is the ordered pair $\left(\frac{10}{3}, -1\right)$. This means that these x and y values are the coordinates of the point where the two lines represented by our equations intersect.
But before we celebrate, it's always wise to double-check our work. We can do this by substituting our x and y values back into both of the original equations to make sure they hold true. This is like checking your answer on a test to avoid silly mistakes. It's a simple step that can save you from a wrong answer!
Let's start with the first equation, $3x + y = 9$:
It checks out! The first equation is satisfied. Now, let's try the second equation, $3x - 5y = 15$:
It checks out again! The second equation is also satisfied. This confirms that our solution is indeed correct. We've successfully navigated the maze of equations and found the hidden treasure!
Comparing to the Multiple-Choice Options and Final Answer
Okay, now that we've confidently solved the system of equations and verified our solution, let's compare it to the multiple-choice options provided. We found that x = $\frac{10}{3}$ and y = -1, so our solution is $\left(\frac{10}{3}, -1\right)$. Now, let's look at the options:
A. $x=\frac{10}{3}, y=-1$
B. $x=\frac{8}{3}, y=1$
C. $x=-1, y=\frac{10}{3}$
D. $x=1, y=\frac{8}{3}$
It's clear that our solution matches option A perfectly! Option A gives us x = $\frac{10}{3}$ and y = -1, which is exactly what we calculated. So, the correct answer to the question βWhat is the solution of $3 x+y=9$ and $3 x-5 y=15$?β is:
A. $x=\frac{10}{3}, y=-1$
We did it! We've successfully solved the system of equations, verified our solution, and identified the correct multiple-choice answer. Give yourselves a pat on the back, guys! You've demonstrated your understanding of the elimination method and your ability to tackle algebraic challenges. Remember, math is like a muscle β the more you exercise it, the stronger it gets. Keep practicing, keep exploring, and you'll continue to grow your problem-solving skills. The world of mathematics is full of fascinating puzzles waiting to be solved, and you're well on your way to becoming a master solver!
Why Other Options Are Incorrect A Deep Dive
To truly master a concept, it's not enough to just know the correct answer; we also need to understand why the other options are incorrect. This deeper understanding solidifies our knowledge and helps us avoid similar mistakes in the future. So, let's take a closer look at options B, C, and D and analyze why they don't work as solutions to our system of equations.
Analyzing Option B
Option B gives us x = $\frac{8}{3}$ and y = 1. To check if this is a solution, we'll substitute these values into our original equations:
Equation 1: $3x + y = 9$
So far, so good! Option B satisfies the first equation. But remember, a solution must satisfy both equations, so let's check the second equation:
Equation 2: $3x - 5y = 15$
Oops! This is not true. Option B does not satisfy the second equation. Therefore, it cannot be the correct solution. This highlights a crucial point: when solving systems of equations, it's essential to check your solution in both equations. Satisfying one equation is not enough; it must satisfy all equations in the system.
Analyzing Option C
Option C gives us x = -1 and y = $\frac{10}{3}$. Let's substitute these values into our equations:
Equation 1: $3x + y = 9$
To add these terms, we need a common denominator. Let's rewrite -3 as $\frac{-9}{3}$:
This is definitely not true! Option C fails to satisfy even the first equation. Therefore, it's incorrect. This example underscores the importance of carefully performing arithmetic operations, especially when dealing with fractions and negative numbers. A small error in arithmetic can lead to a completely wrong answer.
Analyzing Option D
Finally, let's analyze option D, which gives us x = 1 and y = $\frac{8}{3}$. Let's substitute these values into our equations:
Equation 1: $3x + y = 9$
Again, we need a common denominator to add these terms. Let's rewrite 3 as $\frac{9}{3}$:
This is also not true. Option D fails to satisfy the first equation. Therefore, it's incorrect. This reinforces the idea that simply plugging in values without understanding the underlying principles can lead to incorrect conclusions. It's crucial to have a solid understanding of the mathematical concepts and procedures involved.
Key Takeaways from Analyzing Incorrect Options
From our analysis of options B, C, and D, we can draw several key takeaways:
- A solution must satisfy all equations in the system. It's not enough for a solution to work in just one equation; it must work in every equation.
- Careful arithmetic is essential. Even a small error in addition, subtraction, multiplication, or division can lead to a wrong answer.
- Understanding the underlying concepts is crucial. Blindly plugging in values without understanding the math can be misleading. It's important to grasp the principles and processes involved.
- Checking your solution is a must. Always verify your solution by substituting the values back into the original equations. This helps catch errors and builds confidence in your answer.
By understanding why the incorrect options are wrong, we gain a deeper appreciation for the correct solution and the process we used to find it. This is a powerful way to learn and improve our problem-solving skills. Keep asking βwhyβ and digging deeper, guys! Thatβs the spirit of true mathematical exploration.
Mastering Systems of Equations Tips and Tricks
Now that we've thoroughly dissected this problem and understood the solution process, let's zoom out and discuss some general tips and tricks for mastering systems of equations. These strategies will help you tackle a wide range of similar problems with greater confidence and efficiency. Think of these as your secret weapons in the battle against algebraic challenges!
Choosing the Right Method
One of the first decisions you'll need to make when solving a system of equations is which method to use. There are three main methods: graphing, substitution, and elimination. Each method has its strengths and weaknesses, and the best method for a particular problem depends on the specific equations involved.
- Graphing: Graphing involves plotting the equations on a coordinate plane and finding the point of intersection. This method is visually intuitive and can be helpful for understanding the concept of a solution. However, it's not always the most precise method, especially if the solution involves fractions or decimals. Graphing is best suited for simple systems with integer solutions.
- Substitution: Substitution involves solving one equation for one variable and then substituting that expression into the other equation. This method is effective when one of the equations is already solved for a variable or can be easily solved. It's particularly useful when dealing with equations that have a variable with a coefficient of 1.
- Elimination: Elimination, as we've seen, involves manipulating the equations so that when you add or subtract them, one of the variables cancels out. This method is often the most efficient when the coefficients of one of the variables are the same or can be easily made the same by multiplying one or both equations by a constant. It's a powerful technique for solving a wide variety of systems of equations.
The key to choosing the right method is practice and experience. The more problems you solve, the better you'll become at recognizing which method is most suitable for a given situation. It's like choosing the right tool for a job β a screwdriver is great for screws, but you wouldn't use it to hammer a nail!
Spotting Easy Elimination Setups
As we saw in our example problem, the elimination method is particularly straightforward when the coefficients of one of the variables are the same. This allows us to eliminate that variable with a simple subtraction (or addition if the signs are opposite). So, one of the first things you should look for when tackling a system of equations is whether the coefficients of x or y are the same or opposites. If they are, elimination is likely the way to go!
But what if the coefficients aren't the same? Don't worry, you can often make them the same by multiplying one or both equations by a suitable constant. For example, consider the system:
In this case, the coefficients of neither x nor y are the same. However, we can easily make the coefficients of x the same by multiplying the second equation by 2:
Now we have the system:
And we can eliminate x by subtracting the second equation from the first. Spotting these opportunities for easy elimination setups can save you a lot of time and effort!
Dealing with Fractions and Decimals
Systems of equations can sometimes involve fractions or decimals, which can make the arithmetic a bit more challenging. However, there's a simple trick you can use to eliminate fractions or decimals: multiply the entire equation by the least common denominator (LCD) of the fractions or by a power of 10 that will clear the decimals.
For example, consider the equation:
To eliminate the fractions, we can multiply the entire equation by the LCD of 2 and 3, which is 6:
Similarly, if we have an equation with decimals, such as:
We can multiply the entire equation by 10 to clear the decimals:
Clearing fractions or decimals makes the equations easier to work with and reduces the chances of making arithmetic errors.
Checking for Special Cases
Sometimes, systems of equations can have special cases: no solution or infinitely many solutions. It's important to be aware of these possibilities and how to recognize them.
- No Solution: A system of equations has no solution when the lines represented by the equations are parallel. This means they have the same slope but different y-intercepts. Algebraically, this will manifest as a contradiction when you try to solve the system. For example, you might end up with an equation like 0 = 5, which is clearly false.
- Infinitely Many Solutions: A system of equations has infinitely many solutions when the lines represented by the equations are the same. This means they have the same slope and the same y-intercept. Algebraically, this will manifest as an identity when you try to solve the system. For example, you might end up with an equation like 0 = 0, which is always true.
Being mindful of these special cases can prevent you from spending time trying to solve a system that doesn't have a unique solution.
Practice, Practice, Practice!
As with any mathematical skill, the key to mastering systems of equations is practice. The more problems you solve, the more comfortable you'll become with the different methods and techniques. You'll start to recognize patterns, anticipate challenges, and develop your problem-solving intuition. So, don't be afraid to dive in and tackle a variety of problems. Seek out examples in textbooks, online resources, or worksheets, and challenge yourself to solve them. The more you practice, the more confident and proficient you'll become. Remember, every problem you solve is a step closer to mastery! Keep up the awesome work, guys! Math is a journey, not a destination, and the more you explore, the more you'll discover its beauty and power.
What is the solution to the following system of equations: $3x + y = 9$ and $3x - 5y = 15$?
Solve System of Equations 3x+y=9 and 3x-5y=15 A Step-by-Step Guide