Solving Systems Of Equations: A Step-by-Step Guide

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Hey guys! Ever find yourself staring at a system of equations and feeling totally lost? Don't worry, you're not alone! Systems of equations can seem intimidating at first, but with a little know-how, you can totally conquer them. Let's break it down, step by step, using the system you provided as our example. We'll go through it together, making sure you understand each part so you can solve similar problems with confidence. So, grab your pencils, and let's dive in!

Understanding the Problem

Before we jump into solving, let's take a good look at the system of equations we're dealing with:

{x4+y3=14x2+y4=74\begin{cases} \frac{x}{4} + \frac{y}{3} = \frac{1}{4} \\ \frac{x}{2} + \frac{y}{4} = \frac{7}{4} \end{cases}

What exactly is a system of equations? Simply put, it's a set of two or more equations that share the same variables. Our goal here is to find the values of x and y that satisfy both equations simultaneously. Think of it like finding the perfect matching puzzle pieces for both equations at the same time. There are several methods we can use to solve these systems, such as substitution, elimination, or even graphing. But for this example, we'll focus on the elimination method, as it’s often the most straightforward approach for equations like these. Remember, the key is to manipulate the equations in a way that allows us to get rid of one variable, making it easier to solve for the other. Now, let’s get started with clearing those fractions to make our lives easier!

Clearing the Fractions

Fractions can sometimes make equations look a bit messy, right? So, our first step is to clear them out. To do this, we'll multiply each equation by the least common multiple (LCM) of the denominators. This will give us whole numbers, making the equations much easier to work with. For the first equation, x4+y3=14\frac{x}{4} + \frac{y}{3} = \frac{1}{4}, the denominators are 4 and 3. The LCM of 4 and 3 is 12. So, we'll multiply the entire first equation by 12:

12∗(x4+y3)=12∗(14)12 * (\frac{x}{4} + \frac{y}{3}) = 12 * (\frac{1}{4})

This simplifies to:

3x+4y=33x + 4y = 3

Now, let’s tackle the second equation, x2+y4=74\frac{x}{2} + \frac{y}{4} = \frac{7}{4}. Here, the denominators are 2 and 4. The LCM of 2 and 4 is 4. So, we’ll multiply the entire second equation by 4:

4∗(x2+y4)=4∗(74)4 * (\frac{x}{2} + \frac{y}{4}) = 4 * (\frac{7}{4})

This simplifies to:

2x+y=72x + y = 7

See? Much cleaner already! Our system of equations now looks like this:

{3x+4y=32x+y=7\begin{cases} 3x + 4y = 3 \\ 2x + y = 7 \end{cases}

We’ve successfully eliminated the fractions, setting us up for the next step in solving the system: eliminating one of the variables. This is where things start to get really interesting, so let's move on!

Eliminating a Variable

Now that we have our equations with whole numbers, the next strategic move is to eliminate one of the variables. This will leave us with a single equation with just one variable, which is much easier to solve. Looking at our system:

{3x+4y=32x+y=7\begin{cases} 3x + 4y = 3 \\ 2x + y = 7 \end{cases}

We can see that the 'y' variable in the second equation has a coefficient of 1, which makes it a great candidate for elimination. To eliminate 'y', we'll multiply the second equation by -4. This will give us a '-4y' term, which will cancel out the '+4y' term in the first equation when we add the equations together.

So, multiplying the second equation (2x+y=72x + y = 7) by -4, we get:

−4∗(2x+y)=−4∗7-4 * (2x + y) = -4 * 7

Which simplifies to:

−8x−4y=−28-8x - 4y = -28

Now, we can rewrite our system with the modified second equation:

{3x+4y=3−8x−4y=−28\begin{cases} 3x + 4y = 3 \\ -8x - 4y = -28 \end{cases}

Next, we add the two equations together. Notice how the '4y' and '-4y' terms perfectly cancel each other out:

(3x+4y)+(−8x−4y)=3+(−28)(3x + 4y) + (-8x - 4y) = 3 + (-28)

This simplifies to:

−5x=−25-5x = -25

We’ve successfully eliminated 'y' and are left with a simple equation in terms of 'x'. This is a major step forward! In the next section, we'll solve for 'x' and then use that value to find 'y'. Keep going, you're doing great!

Solving for x

Alright, we've made excellent progress! We've eliminated 'y' and now we have a straightforward equation to solve for 'x':

−5x=−25-5x = -25

To isolate 'x', we simply divide both sides of the equation by -5:

−5x−5=−25−5\frac{-5x}{-5} = \frac{-25}{-5}

This gives us:

x=5x = 5

Woohoo! We've found the value of 'x'. Now we know that one piece of our puzzle is x = 5. But remember, we're solving a system of equations, so we need to find the value of 'y' as well. Don't worry, we're more than halfway there! With the value of 'x' in hand, we can now substitute it back into one of our original equations to solve for 'y'. This is the next crucial step, so let's move on to finding 'y'. You’re doing fantastic – keep up the momentum!

Solving for y

Now that we've successfully found the value of x, which is 5, it's time to find the value of y. To do this, we'll substitute the value of x back into one of our original equations. We can choose either equation, but let's pick the second one because it looks a bit simpler:

2x+y=72x + y = 7

Substitute x=5x = 5 into the equation:

2(5)+y=72(5) + y = 7

This simplifies to:

10+y=710 + y = 7

Now, to isolate y, we subtract 10 from both sides of the equation:

y=7−10y = 7 - 10

Which gives us:

y=−3y = -3

Excellent! We've found the value of y. So, we now know that y=−3y = -3. We're almost at the finish line! We've found the values of both x and y, but it's always a good idea to double-check our solution to make sure it's correct. In the next section, we'll verify our solution by plugging the values of x and y back into both original equations. Let's make sure everything checks out!

Verifying the Solution

We've arrived at a potential solution: x=5x = 5 and y=−3y = -3. But before we celebrate, it's crucial to verify that these values actually satisfy both of our original equations. This step ensures we haven't made any mistakes along the way.

Let's start with the first original equation:

x4+y3=14\frac{x}{4} + \frac{y}{3} = \frac{1}{4}

Substitute x=5x = 5 and y=−3y = -3 into the equation:

54+−33=14\frac{5}{4} + \frac{-3}{3} = \frac{1}{4}

Simplify the equation:

54−1=14\frac{5}{4} - 1 = \frac{1}{4}

To combine the terms on the left side, we need a common denominator. We can rewrite 1 as 44\frac{4}{4}:

54−44=14\frac{5}{4} - \frac{4}{4} = \frac{1}{4}

Which simplifies to:

14=14\frac{1}{4} = \frac{1}{4}

Great! The values satisfy the first equation. Now, let's check the second original equation:

x2+y4=74\frac{x}{2} + \frac{y}{4} = \frac{7}{4}

Substitute x=5x = 5 and y=−3y = -3 into the equation:

52+−34=74\frac{5}{2} + \frac{-3}{4} = \frac{7}{4}

To combine the terms on the left side, we need a common denominator. The LCM of 2 and 4 is 4, so we rewrite 52\frac{5}{2} as 104\frac{10}{4}:

104−34=74\frac{10}{4} - \frac{3}{4} = \frac{7}{4}

Which simplifies to:

74=74\frac{7}{4} = \frac{7}{4}

Fantastic! The values also satisfy the second equation. This confirms that our solution is correct. We've successfully solved the system of equations!

Final Answer

We did it! After working through all the steps, we've found the solution to the system of equations:

{x4+y3=14x2+y4=74\begin{cases} \frac{x}{4} + \frac{y}{3} = \frac{1}{4} \\ \frac{x}{2} + \frac{y}{4} = \frac{7}{4} \end{cases}

The solution is:

x=5x = 5 and y=−3y = -3

So, the values that satisfy both equations simultaneously are x=5x = 5 and y=−3y = -3.

Congratulations! You've successfully navigated through the process of solving a system of equations. Remember, the key is to take it step by step, stay organized, and double-check your work. With practice, you'll become a pro at solving these types of problems. Keep up the great work, and don't hesitate to tackle more challenges!