10 Solved Examples: First-Degree System Of Equations

Oct 10, 2025 by ADMIN 53 views

Hey guys! Ever get tripped up by systems of equations? Don't worry, you're not alone! First-degree systems of equations can seem intimidating, but once you break them down, they're totally manageable. This article is all about demystifying these problems. We're diving into ten solved examples that will walk you through the process step-by-step. Get ready to level up your math skills!

What are First-Degree Systems of Equations?

Before we jump into the examples, let's make sure we're all on the same page. A first-degree system of equations, also known as a linear system, is a set of two or more equations containing two or more variables, where the highest power of the variables is 1. These systems are fundamental in mathematics and have wide-ranging applications in fields like physics, engineering, economics, and computer science. Understanding how to solve them opens doors to solving real-world problems. Imagine calculating the break-even point for a business, determining the optimal mix of ingredients for a recipe, or even modeling the flow of traffic – these are just a few examples of where systems of equations come into play.

The core idea behind solving a system of equations is to find values for the variables that satisfy all equations simultaneously. This means that when you substitute the values you find back into the original equations, both sides of each equation will be equal. There are several methods for solving these systems, each with its own strengths and weaknesses, such as substitution, elimination (also called addition), and graphing. The best method to use often depends on the specific equations in the system. For instance, if one equation is already solved for one variable in terms of the other, substitution might be the most straightforward approach. On the other hand, if the coefficients of one variable in the equations are opposites or easy multiples of each other, elimination could be more efficient. Graphing provides a visual representation of the solution, which is the point where the lines representing the equations intersect. As we go through the examples, we'll highlight different techniques and when they might be most useful. So, let's dive in and conquer these equations together!

Methods for Solving First-Degree Systems

There are several methods available to tackle first-degree systems of equations, each with its own advantages depending on the structure of the equations. Mastering these methods provides a versatile toolkit for solving various problems. Let's explore the most common techniques:

Substitution Method: This method involves solving one equation for one variable and then substituting that expression into the other equation. This reduces the system to a single equation with one variable, which can then be solved directly. Once you find the value of one variable, you can substitute it back into either of the original equations to find the value of the other variable. The substitution method is particularly useful when one of the equations is already solved for a variable or can be easily rearranged to do so. It's a powerful technique for simplifying complex systems into manageable parts.
Elimination Method (Addition Method): The elimination method, also known as the addition method, focuses on eliminating one of the variables by adding the equations together. This often requires multiplying one or both equations by a constant so that the coefficients of one variable are opposites. When the equations are added, that variable cancels out, leaving you with a single equation in one variable. Like the substitution method, this simplifies the problem, making it easier to solve. This method is especially effective when the coefficients of one variable are already opposites or can be easily made so. It's a great way to streamline the solving process.
Graphing Method: This method provides a visual approach to solving systems of equations. Each equation is graphed on a coordinate plane, and the solution to the system is the point where the lines intersect. Graphing can be a helpful way to understand the concept of a solution as a point that satisfies both equations simultaneously. It's also useful for systems with two variables and provides a clear visual representation of the problem. However, it may not be the most precise method for finding solutions, especially if the intersection point has non-integer coordinates. Despite this, graphing offers valuable insight into the nature of the solutions and the relationship between the equations.

Each method has its strengths, and the choice of which one to use often depends on the specific problem. As we work through the examples, we'll see how these methods are applied in different situations, giving you a better understanding of when each one is most effective. So, let's get started with our examples and put these methods into action!

Example 1: Solving by Substitution

Let’s kick things off with a classic example using the substitution method. Consider the following system of equations:

x + y = 5
2x - y = 1

Our goal is to find values for x and y that satisfy both equations. The substitution method is perfect here because we can easily solve the first equation for one variable in terms of the other. Let's solve the first equation for y:

y = 5 - x

Now, we'll substitute this expression for y into the second equation:

2x - (5 - x) = 1

See how we've replaced y with 5 - x? This gives us a single equation with just x as the variable. Let's simplify and solve for x:

2x - 5 + x = 1
3x - 5 = 1
3x = 6
x = 2

Great! We've found that x = 2. Now we can substitute this value back into our expression for y:

y = 5 - x
y = 5 - 2
y = 3

So, we've found that y = 3. Our solution is x = 2 and y = 3. To check our answer, we can substitute these values back into the original equations:

2 + 3 = 5  (True)
2(2) - 3 = 1 (True)

Both equations hold true, so our solution is correct! The substitution method allowed us to break down the system by expressing one variable in terms of the other, making the problem much easier to solve. In this case, it was particularly efficient because the first equation could be easily rearranged. Keep this example in mind as we move on to more complex scenarios.

Example 2: Solving by Elimination

Now, let's tackle a system of equations using the elimination method. This technique is also super useful, especially when the coefficients of one of the variables are easy to manipulate. Check out this system:

3x + 2y = 7
4x - 2y = 0

Notice anything special about the y terms? They have opposite coefficients! This makes the elimination method a fantastic choice. All we need to do is add the two equations together:

(3x + 2y) + (4x - 2y) = 7 + 0

When we add them, the 2y and -2y terms cancel each other out:

7x = 7

Now we have a simple equation to solve for x:

x = 1

Awesome! We've found that x = 1. To find y, we can substitute this value into either of the original equations. Let's use the first one:

3(1) + 2y = 7
3 + 2y = 7
2y = 4
y = 2

So, we've got y = 2. Our solution is x = 1 and y = 2. Let's verify this by plugging these values back into the original equations:

3(1) + 2(2) = 7  (True)
4(1) - 2(2) = 0  (True)

Both equations are satisfied, so our solution is spot on! The elimination method was incredibly efficient in this example because the y terms had opposite coefficients. This allowed us to eliminate one variable with a simple addition, making the problem much easier to solve. This highlights the importance of recognizing the structure of the equations to choose the most effective method.

Example 3: Elimination with Multiplication

Sometimes, the equations aren't quite set up perfectly for direct elimination, but with a little manipulation, we can make it work! Consider this system:

2x + 3y = 8
x + 2y = 5

The coefficients of x and y aren't opposites, and we can't directly add the equations to eliminate a variable. But no worries, we can multiply one of the equations by a constant to create opposite coefficients. Let's eliminate x. We can multiply the second equation by -2:

-2(x + 2y) = -2(5)
-2x - 4y = -10

Now we have a new system:

2x + 3y = 8
-2x - 4y = -10

Look at that! The x coefficients are now opposites. We can add the equations together:

(2x + 3y) + (-2x - 4y) = 8 + (-10)

This simplifies to:

-y = -2

Solving for y, we get:

y = 2

Fantastic! We've found y = 2. Now, let's substitute this value back into one of the original equations. We'll use the second one:

x + 2(2) = 5
x + 4 = 5
x = 1

So, x = 1. Our solution is x = 1 and y = 2. Let's check our work by plugging these values into the original equations:

2(1) + 3(2) = 8  (True)
1 + 2(2) = 5  (True)

Both equations are satisfied, so we've nailed it! This example demonstrates how multiplying one or both equations by a constant can be a crucial step in the elimination method. By strategically manipulating the equations, we can create opportunities to eliminate variables and simplify the system.

Example 4: Substitution with Fractions

Fractions can sometimes make systems of equations look scary, but don't let them intimidate you! The techniques we've already learned still apply. Let's tackle this system:

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

To make things a bit easier, let's first get rid of the fractions. We can do this by multiplying each equation by the least common multiple (LCM) of the denominators. For the first equation, the LCM of 2 is simply 2. For the second equation, the LCM of 3 is 3. So, let's multiply:

2(x/2 + y) = 2(4)  =>  x + 2y = 8
3(x - y/3) = 3(1)  =>  3x - y = 3

Now we have a much cleaner system:

x + 2y = 8
3x - y = 3

The substitution method seems like a good approach here. Let's solve the first equation for x:

x = 8 - 2y

Now we can substitute this expression for x into the second equation:

3(8 - 2y) - y = 3

Simplify and solve for y:

24 - 6y - y = 3
24 - 7y = 3
-7y = -21
y = 3

Excellent! We've found y = 3. Let's substitute this back into our expression for x:

x = 8 - 2(3)
x = 8 - 6
x = 2

So, x = 2. Our solution is x = 2 and y = 3. Let's check our solution in the original equations (with fractions):

2/2 + 3 = 4  (True)
2 - 3/3 = 1  (True)

Both equations are satisfied! This example highlights a useful strategy for dealing with fractions in systems of equations: clear the fractions by multiplying each equation by the LCM of the denominators. This often simplifies the system and makes it easier to solve using standard methods.

Example 5: Elimination with Decimals

Just like fractions, decimals can sometimes make systems of equations look a bit intimidating. But don't worry, we can handle them! Let's look at this system:

0.2x + 0.3y = 1.7
0.5x - 0.2y = 1.1

The key to dealing with decimals is to eliminate them by multiplying each equation by a power of 10. In this case, the highest number of decimal places is one, so we can multiply both equations by 10:

10(0.2x + 0.3y) = 10(1.7)  =>  2x + 3y = 17
10(0.5x - 0.2y) = 10(1.1)  =>  5x - 2y = 11

Now we have a system with whole numbers:

2x + 3y = 17
5x - 2y = 11

This looks like a good candidate for the elimination method. Let's eliminate y. To do this, we need to make the coefficients of y opposites. We can multiply the first equation by 2 and the second equation by 3:

2(2x + 3y) = 2(17)  =>  4x + 6y = 34
3(5x - 2y) = 3(11)  =>  15x - 6y = 33

Now our system is:

4x + 6y = 34
15x - 6y = 33

The y terms have opposite coefficients, so we can add the equations:

(4x + 6y) + (15x - 6y) = 34 + 33

This simplifies to:

19x = 67

Solving for x, we get:

x = 67/19

Okay, x is a fraction, but that's perfectly fine! Now, let's substitute this value back into one of our equations. We'll use the first one with whole numbers:

2(67/19) + 3y = 17

Solving for y:

134/19 + 3y = 17
3y = 17 - 134/19
3y = (323 - 134)/19
3y = 189/19
y = 63/19

So, y = 63/19. Our solution is x = 67/19 and y = 63/19. It's always a good idea to check our answer, especially with fractions. Let's plug these values back into the original equations with decimals (using a calculator for the arithmetic):

0.  2(67/19) + 0.3(63/19) = 1.7  (True)
1.  5(67/19) - 0.2(63/19) = 1.1  (True)

Both equations hold true! This example demonstrates the key strategy for dealing with decimals in systems of equations: clear the decimals by multiplying each equation by the appropriate power of 10. This simplifies the system and allows us to use standard methods to find the solution.

Example 6: Systems with No Solution

Sometimes, systems of equations don't have a solution. It might seem strange, but it's important to recognize these cases. Let's take a look at this system:

x + y = 3
2x + 2y = 8

Let's try the elimination method. We can multiply the first equation by -2:

-2(x + y) = -2(3)
-2x - 2y = -6

Now our system is:

-2x - 2y = -6
2x + 2y = 8

Add the equations together:

(-2x - 2y) + (2x + 2y) = -6 + 8
0 = 2

Wait a minute... 0 = 2? That's definitely not true! This contradiction tells us that this system has no solution. What does this mean geometrically? If we were to graph these equations, we would find that they are parallel lines. Parallel lines never intersect, which means there's no point (x, y) that satisfies both equations simultaneously.

Example 7: Systems with Infinite Solutions

On the flip side, some systems have not just one solution, but infinitely many! Let's see how this happens with the following system:

x - y = 1
2x - 2y = 2

Let's try the elimination method again. We can multiply the first equation by -2:

-2(x - y) = -2(1)
-2x + 2y = -2

Now our system is:

-2x + 2y = -2
2x - 2y = 2

Add the equations together:

(-2x + 2y) + (2x - 2y) = -2 + 2
0 = 0

This time, we got 0 = 0, which is always true. This means that the two equations are actually dependent. In other words, they represent the same line. Any point on that line will satisfy both equations, so there are infinitely many solutions. When we graph these equations, we'll see that they overlap completely.

Example 8: A Word Problem

Let's see how systems of equations can help us solve real-world problems. Imagine this scenario: A movie theater sells tickets for $10 for adults and $6 for children. If a total of 200 tickets were sold and the revenue was $1440, how many adult and child tickets were sold?

First, let's define our variables:

Let a be the number of adult tickets sold.
Let c be the number of child tickets sold.

Now we can set up our system of equations based on the information given:

a + c = 200  (Total number of tickets)
10a + 6c = 1440 (Total revenue)

Let's use the substitution method. Solve the first equation for a:

a = 200 - c

Substitute this into the second equation:

10(200 - c) + 6c = 1440

Simplify and solve for c:

2000 - 10c + 6c = 1440
2000 - 4c = 1440
-4c = -560
c = 140

So, 140 child tickets were sold. Now let's find the number of adult tickets:

a = 200 - c
a = 200 - 140
a = 60

Therefore, 60 adult tickets were sold. We can check our answer by plugging these values back into our original equations:

60 + 140 = 200 (True)
10(60) + 6(140) = 600 + 840 = 1440 (True)

Our solution checks out! We sold 60 adult tickets and 140 child tickets.

Example 9: Another Word Problem

Here's another word problem to flex our system-solving muscles: A boat travels 60 miles downstream in 3 hours and 60 miles upstream in 5 hours. Find the speed of the boat in still water and the speed of the current.

Let's define our variables:

Let b be the speed of the boat in still water (in miles per hour).
Let c be the speed of the current (in miles per hour).

When the boat travels downstream, the current helps it along, so the effective speed is b + c. When the boat travels upstream, the current works against it, so the effective speed is b - c. We can use the formula distance = speed × time to set up our equations:

3(b + c) = 60  (Downstream)
5(b - c) = 60  (Upstream)

Let's simplify these equations by dividing both sides of the first equation by 3 and both sides of the second equation by 5:

b + c = 20
b - c = 12

The elimination method is perfect here! We can add the equations together:

(b + c) + (b - c) = 20 + 12
2b = 32
b = 16

So, the speed of the boat in still water is 16 mph. Now let's find the speed of the current by substituting b = 16 into one of the equations. We'll use the first one:

16 + c = 20
c = 4

Therefore, the speed of the current is 4 mph. We can check our answer by plugging these values back into our original equations:

3(16 + 4) = 3(20) = 60 (True)
5(16 - 4) = 5(12) = 60 (True)

Our solution is correct! The boat's speed in still water is 16 mph, and the current's speed is 4 mph.

Example 10: A Mixture Problem

Here's a classic mixture problem: A chemist has two solutions of sulfuric acid. One is 20% sulfuric acid, and the other is 40% sulfuric acid. How many liters of each solution should the chemist mix to obtain 100 liters of a 32% sulfuric acid solution?

Let's define our variables:

Let x be the number of liters of the 20% solution.
Let y be the number of liters of the 40% solution.

We can set up two equations based on the given information. The first equation represents the total volume of the mixture:

x + y = 100

The second equation represents the amount of sulfuric acid in the mixture:

0.20x + 0.40y = 0.32(100)

Let's simplify the second equation:

0.  20x + 0.40y = 32

To make things easier, we can multiply this equation by 100 to eliminate the decimals:

20x + 40y = 3200

Now we have our system:

x + y = 100
20x + 40y = 3200

Let's use the substitution method. Solve the first equation for x:

x = 100 - y

Substitute this into the second equation:

20(100 - y) + 40y = 3200

Simplify and solve for y:

2000 - 20y + 40y = 3200
20y = 1200
y = 60

So, the chemist needs 60 liters of the 40% solution. Now let's find the amount of the 20% solution:

x = 100 - y
x = 100 - 60
x = 40

Therefore, the chemist needs 40 liters of the 20% solution. Let's check our answer:

40 + 60 = 100 (True)
0.  20(40) + 0.40(60) = 8 + 24 = 32 (True)

Our solution is correct! The chemist should mix 40 liters of the 20% solution and 60 liters of the 40% solution.

Conclusion

So there you have it – ten solved examples of first-degree systems of equations! We've covered various methods like substitution and elimination, dealt with fractions and decimals, and even tackled word problems. The key to mastering these systems is practice, practice, practice! Remember, there's often more than one way to solve a problem, so find the method that clicks with you. Keep these examples handy as you work through your own problems, and you'll be solving systems of equations like a pro in no time. Keep up the great work, guys!