Solve Simultaneous Equations For X, Y, And A

Hey guys! Ready to tackle some math challenges that will really get your brain buzzing? Today, we're diving deep into the world of simultaneous equations. You know, those nifty problems where you have multiple equations and you need to find the values of multiple variables that make all of them true at the same time. It's like cracking a code, and we're going to break it down step-by-step. Our mission, should we choose to accept it, is to find the values of x, y, and a by solving the following set of equations:

6x - 7y = -10
12x - 5y = 16
2x + ay = 10

Don't let the 'a' throw you off; we'll get to that part later. First things first, we need to nail down the values of 'x' and 'y' using the first two equations. These two equations only involve 'x' and 'y', making them our starting point. We'll be using a technique called the elimination method here. It's a super effective way to solve systems of equations by strategically adding or subtracting equations to eliminate one of the variables. So, grab your notebooks, sharpen those pencils, and let's get this math party started!

Finding x and y: The Foundation

Alright, let's focus on our first two equations, the ones that will give us the bedrock values of x and y:

Equation 1: 6x - 7y = -10 Equation 2: 12x - 5y = 16

Our goal with the elimination method is to make the coefficients of either 'x' or 'y' in both equations opposites, so when we add the equations together, one variable vanishes. Looking at the 'x' coefficients, we have 6 and 12. If we multiply Equation 1 by -2, the 'x' coefficient becomes -12, which is the opposite of 12 in Equation 2. This is exactly what we want!

Let's do that:

Multiply Equation 1 by -2: -2 * (6x - 7y) = -2 * (-10) This gives us: -12x + 14y = 20 (Let's call this Equation 3)

Now, we have our modified Equation 3 and our original Equation 2:

Equation 3: -12x + 14y = 20 Equation 2: 12x - 5y = 16

See how the 'x' terms have opposite signs? Perfect! Now, let's add Equation 3 and Equation 2 together:

(-12x + 14y) + (12x - 5y) = 20 + 16

The '-12x' and '+12x' cancel each other out (poof!), leaving us with:

14y - 5y = 36

Combine the 'y' terms:

9y = 36

To find 'y', we just need to divide both sides by 9:

y = 36 / 9

y = 4

Boom! We've found our first variable, y = 4. This is a huge step, guys. With 'y' in hand, we can now substitute this value back into one of our original equations (Equation 1 or Equation 2) to solve for 'x'. Let's use Equation 1 because the numbers look a bit friendlier.

Substitute y = 4 into Equation 1 (6x - 7y = -10):

6x - 7(4) = -10

6x - 28 = -10

Now, we want to isolate the 'x' term. Add 28 to both sides of the equation:

6x = -10 + 28

6x = 18

Finally, divide both sides by 6 to get the value of 'x':

x = 18 / 6

x = 3

There we have it! x = 3 and y = 4. We've successfully solved the first part of our puzzle. Before we move on to finding 'a', it's always a good idea to quickly check these values in the other original equation (Equation 2) to make sure everything lines up. Let's do that:

Check with Equation 2 (12x - 5y = 16): Substitute x = 3 and y = 4: 12(3) - 5(4) = 16 36 - 20 = 16 16 = 16

It checks out perfectly! So, we're confident that x = 3 and y = 4 are the correct values. High five!

Incorporating the Third Equation: Finding 'a'

Now that we've cracked the code for 'x' and 'y', it's time to bring in our third equation and figure out the value of a. The third equation is:

2x + ay = 10

We already know that x = 3 and y = 4. We can substitute these known values into this equation. This will leave us with an equation that only contains our unknown, 'a'. This is the beauty of solving these systems step-by-step; each solved variable simplifies the next step.

Substitute x = 3 and y = 4 into 2x + ay = 10:

2(3) + a(4) = 10

6 + 4a = 10

Our goal now is to isolate 'a'. First, subtract 6 from both sides of the equation:

4a = 10 - 6

4a = 4

And to find 'a', we divide both sides by 4:

a = 4 / 4

a = 1

And there you have it, folks! We've found the value of a = 1. So, the complete solution to this system of simultaneous equations is x = 3, y = 4, and a = 1. It's always satisfying when all the pieces fit together, right?

Final Check and Recap

Before we wrap this up, let's just do one final quick check to ensure our value of 'a' is correct by plugging x=3, y=4, and a=1 back into the third equation 2x + ay = 10.

2(3) + (1)(4) = 10 6 + 4 = 10 10 = 10

It holds true! This confirms our calculations are solid.

So, to recap, we tackled a system of three simultaneous equations involving variables x, y, and a. We started by using the two equations with only x and y to find their values. We employed the elimination method, which involved multiplying one equation to create opposite coefficients, allowing us to eliminate 'x' and solve for 'y'. Once 'y' was found, we substituted it back into one of the original equations to find 'x'. Finally, with the values of 'x' and 'y' secured, we substituted them into the third equation to solve for 'a'.

This problem is a classic example of how algebraic manipulation, specifically using methods like elimination (or substitution, which is another great way to solve these!), allows us to systematically unravel complex mathematical problems. Remember, practice is key, guys! The more you work through these types of problems, the more intuitive they become. Keep those math skills sharp, and don't shy away from a good challenge. Until next time, happy solving!