Solving 5x + 4y = -30 And 3x - 9y = -18 A Step By Step Guide

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Hey guys! Today, we're diving into the exciting world of solving systems of equations. Specifically, we're going to tackle the system:

{5x+4y=−303x−9y=−18\begin{cases} 5x + 4y = -30 \\ 3x - 9y = -18 \end{cases}

This might look a bit intimidating at first, but don't worry! We'll break it down step-by-step and you'll see it's totally manageable. We'll explore different methods to solve this, ensuring you grasp the concepts thoroughly. So, let's get started and unlock the secrets to solving this system of equations!

Understanding Systems of Equations

Before we jump into solving, let's quickly recap what a system of equations actually is. In simple terms, a system of equations is a set of two or more equations that involve the same variables. Our goal is to find the values of these variables that satisfy all equations in the system simultaneously. Think of it like finding the perfect combination that makes everything click into place. Systems of equations pop up everywhere in math, science, and even everyday life, so mastering them is super valuable.

Why are systems of equations so important? Well, they allow us to model and solve problems that involve multiple unknowns and relationships. For example, you might use a system of equations to determine the break-even point for a business, calculate the trajectory of a projectile, or even figure out the nutritional content of a meal. The applications are vast and varied! So, understanding how to solve them opens up a whole new world of problem-solving possibilities. In our specific example, we have two linear equations with two variables, x and y. This means we're looking for a single pair of values (x, y) that makes both equations true at the same time. There are several ways to find this magical pair, and we'll explore the most common methods in detail.

Methods to Solve Systems of Equations

There are primarily two main methods we can use to solve systems of equations like this: the substitution method and the elimination method. Each method has its own strengths, and the best choice often depends on the specific system you're dealing with. Let's take a closer look at each one.

1. The Substitution Method

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This effectively reduces the system to a single equation with a single variable, which we can then solve easily. Once we've found the value of one variable, we can substitute it back into either of the original equations to find the value of the other variable. It's like a clever game of swapping and simplifying!

Here's a step-by-step breakdown of how the substitution method works:

  1. Solve one equation for one variable: Choose the equation that looks easiest to manipulate. In our system, the second equation (3x - 9y = -18) might be a good candidate since all the coefficients are divisible by 3. We could solve this equation for x or y, but solving for x seems a bit simpler in this case. Divide the entire equation by 3: x - 3y = -6. Then, add 3y to both sides: x = 3y - 6.
  2. Substitute: Now, we substitute this expression for x (3y - 6) into the other equation (5x + 4y = -30). This gives us: 5(3y - 6) + 4y = -30. Notice that we've successfully eliminated x and now we have an equation with only y.
  3. Solve for the remaining variable: Simplify and solve the equation for y. Distribute the 5: 15y - 30 + 4y = -30. Combine like terms: 19y - 30 = -30. Add 30 to both sides: 19y = 0. Finally, divide by 19: y = 0.
  4. Substitute back: We've found that y = 0. Now we substitute this value back into either of the original equations (or the expression we found for x) to find x. Let's use the expression x = 3y - 6: x = 3(0) - 6. Simplify: x = -6.
  5. Check your solution: It's always a good idea to check your solution by substituting both x and y values into both original equations to make sure they hold true. This helps prevent silly mistakes!

2. The Elimination Method

The elimination method, also known as the addition method, involves manipulating the equations so that the coefficients of one of the variables are opposites (e.g., 2 and -2). Then, we add the equations together, which eliminates that variable, leaving us with a single equation with one variable. Just like in substitution, we can then solve for the remaining variable and substitute back to find the other one. Think of it as a strategic game of cancelling out terms!

Here's the step-by-step process for the elimination method:

  1. Multiply equations (if needed): Our goal is to make the coefficients of either x or y opposites. Looking at our system,

    {5x+4y=−303x−9y=−18\begin{cases} 5x + 4y = -30 \\ 3x - 9y = -18 \end{cases}

    it might be easiest to eliminate x. To do this, we can multiply the first equation by -3 and the second equation by 5. This will give us -15x in the first equation and 15x in the second equation.

    • Multiply the first equation by -3: -3(5x + 4y) = -3(-30) --> -15x - 12y = 90
    • Multiply the second equation by 5: 5(3x - 9y) = 5(-18) --> 15x - 45y = -90

  2. Add the equations: Now, we add the two modified equations together:

    −15x−12y=9015x−45y=−90−57y=0\begin{aligned} -15x - 12y &= 90 \\ 15x - 45y &= -90 \\ \hline -57y &= 0 \end{aligned}

    Notice how the x terms cancel out, leaving us with an equation in just y.

  3. Solve for the remaining variable: Solve the equation -57y = 0 for y. Divide both sides by -57: y = 0.

  4. Substitute back: Now that we know y = 0, we substitute this value back into either of the original equations to solve for x. Let's use the first equation, 5x + 4y = -30: 5x + 4(0) = -30. Simplify: 5x = -30. Divide both sides by 5: x = -6.

  5. Check your solution: Again, it's crucial to check our solution by plugging the values of x and y back into both original equations. This confirms that our solution is correct.

Solving the System Using Both Methods

Let's put our knowledge to the test and solve the given system of equations using both the substitution and elimination methods. This will not only solidify our understanding but also demonstrate how both methods lead to the same solution.

1. Solving with the Substitution Method

As we discussed earlier, the substitution method involves solving one equation for one variable and substituting that expression into the other equation. Looking at our system:

{5x+4y=−303x−9y=−18\begin{cases} 5x + 4y = -30 \\ 3x - 9y = -18 \end{cases}

The second equation, 3x - 9y = -18, seems easier to manipulate. Let's solve it for x:

  1. Divide the second equation by 3: x - 3y = -6
  2. Add 3y to both sides: x = 3y - 6

Now we substitute this expression for x into the first equation:

5(3y - 6) + 4y = -30

Next, we solve for y:

  1. Distribute the 5: 15y - 30 + 4y = -30
  2. Combine like terms: 19y - 30 = -30
  3. Add 30 to both sides: 19y = 0
  4. Divide by 19: y = 0

We've found that y = 0. Now we substitute this value back into the expression x = 3y - 6 to find x:

x = 3(0) - 6 x = -6

So, using the substitution method, we find that x = -6 and y = 0.

2. Solving with the Elimination Method

Now, let's tackle the same system using the elimination method. Remember, this method involves manipulating the equations so that the coefficients of one of the variables are opposites, and then adding the equations together.

{5x+4y=−303x−9y=−18\begin{cases} 5x + 4y = -30 \\ 3x - 9y = -18 \end{cases}

To eliminate x, we can multiply the first equation by -3 and the second equation by 5:

  • Multiply the first equation by -3: -15x - 12y = 90
  • Multiply the second equation by 5: 15x - 45y = -90

Now, add the two equations together:

−15x−12y=9015x−45y=−90−57y=0\begin{aligned} -15x - 12y &= 90 \\ 15x - 45y &= -90 \\ \hline -57y &= 0 \end{aligned}

Solve for y:

-57y = 0 y = 0

Now that we have y = 0, substitute it back into either of the original equations to solve for x. Let's use the first equation, 5x + 4y = -30:

5x + 4(0) = -30 5x = -30 x = -6

Using the elimination method, we also arrive at the solution x = -6 and y = 0.

The Solution and Verification

As we've demonstrated using both the substitution and elimination methods, the solution to the system of equations is x = -6 and y = 0. This means that the point (-6, 0) is the intersection of the two lines represented by these equations. It's the one and only point that satisfies both equations simultaneously. Pretty cool, huh?

But we're not done yet! It's crucial to verify our solution to ensure we haven't made any sneaky errors along the way. To do this, we simply substitute our values for x and y back into both original equations and check if they hold true.

Let's start with the first equation, 5x + 4y = -30:

5(-6) + 4(0) = -30 -30 + 0 = -30 -30 = -30

Awesome! The first equation checks out.

Now, let's check the second equation, 3x - 9y = -18:

3(-6) - 9(0) = -18 -18 - 0 = -18 -18 = -18

Fantastic! The second equation also holds true. Since our solution satisfies both equations, we can confidently say that x = -6 and y = 0 is indeed the correct solution to the system.

Conclusion

So there you have it, guys! We've successfully solved the system of equations:

{5x+4y=−303x−9y=−18\begin{cases} 5x + 4y = -30 \\ 3x - 9y = -18 \end{cases}

We explored two powerful methods – substitution and elimination – and saw how both lead us to the same solution: x = -6 and y = 0. We also emphasized the importance of verifying our solution to ensure accuracy.

Solving systems of equations is a fundamental skill in mathematics, and it's something you'll encounter in various contexts throughout your studies and beyond. Mastering these methods will not only boost your problem-solving abilities but also give you a deeper appreciation for the elegance and power of mathematical tools. So, keep practicing, keep exploring, and don't be afraid to tackle those tricky equations! You've got this!