Solve For X: 2(x-3)+9=3(x+1)+x

by Andrew McMorgan 31 views

Hey guys! Let's dive into this equation and figure out what xx is. Sometimes math problems look a little intimidating, but trust me, once you break them down, they're totally manageable. Today, we're tackling the equation 2(xβˆ’3)+9=3(x+1)+x2(x-3)+9=3(x+1)+x. The goal here is to isolate xx and find its value. We've got some options to choose from: x=βˆ’3x=-3, x=βˆ’1x=-1, x=0x=0, or x=3x=3. Let's get this solved!

Step-by-Step Solution to Find the Value of x

Alright, let's get down to business and solve this equation step-by-step. Our mission is to find the value of x in the given equation 2(xβˆ’3)+9=3(x+1)+x2(x-3)+9=3(x+1)+x. Remember, the key to solving algebraic equations is to simplify both sides first and then manipulate the equation to get xx all by itself on one side. This might seem like a trek, but we'll take it one move at a time, and you'll see that solving for x can be pretty straightforward. So, grab your mental calculators, and let's get cracking!

Simplifying the Left Side of the Equation

First up, let's focus on the left side of our equation: 2(xβˆ’3)+92(x-3)+9. The first thing we need to do here is to distribute the 22 to both terms inside the parentheses. This means multiplying 22 by xx and then multiplying 22 by βˆ’3-3. So, 22 times xx gives us 2x2x, and 22 times βˆ’3-3 gives us βˆ’6-6. Now, our expression looks like 2xβˆ’6+92x - 6 + 9. We can simplify this further by combining the constant terms, βˆ’6-6 and +9+9. Adding these together, we get +3+3. So, the simplified left side of the equation is 2x+32x + 3. Keeping this simplified form in mind is crucial because it makes the rest of the solving process much cleaner. We’ve successfully simplified the left side, and now we're ready to tackle the other half of the equation. It’s all about breaking down the problem into smaller, manageable pieces, and this step is a big one!

Simplifying the Right Side of the Equation

Now, let's pivot to the right side of our equation: 3(x+1)+x3(x+1)+x. Similar to the left side, we need to start by distributing the 33 to the terms inside the parentheses. So, 33 times xx gives us 3x3x, and 33 times 11 gives us 33. This part of the expression now becomes 3x+3+x3x + 3 + x. See that extra xx at the end? We can combine it with the 3x3x. Think of it as 3x+1x3x + 1x. Adding these together, we get 4x4x. So, the simplified right side of the equation is 4x+34x + 3. We've now simplified the right side, and it looks much neater. This simplification is super important for the next steps in solving for x. We're getting closer to finding that mystery value!

Equating the Simplified Sides

We've done the hard work of simplifying both sides of the equation. On the left, we have 2x+32x + 3, and on the right, we have 4x+34x + 3. Now, we set these two simplified expressions equal to each other: 2x+3=4x+32x + 3 = 4x + 3. This is the core equation we need to solve for xx. The goal now is to get all the terms with xx on one side and all the constant terms on the other. This step is where we start to really isolate x. By having both sides simplified, it’s much easier to see how to move the terms around. Remember, whatever operation you do to one side of the equation, you must do to the other side to keep it balanced. We’re moving steadily towards our answer, guys!

Isolating the Variable x

We're at the stage where we have 2x+3=4x+32x + 3 = 4x + 3, and our next big move is to isolate the variable x. Let's start by getting all the xx terms together. A good strategy is to subtract 2x2x from both sides of the equation. Why? Because it will leave us with xx terms only on the right side, and it's generally easier to work with positive coefficients for xx. So, subtracting 2x2x from both sides gives us: (2x+3)βˆ’2x=(4x+3)βˆ’2x(2x + 3) - 2x = (4x + 3) - 2x. On the left side, 2xβˆ’2x2x - 2x cancels out, leaving us with just 33. On the right side, 4xβˆ’2x4x - 2x equals 2x2x, so we have 3+2x3 + 2x. Our equation now looks like 3=3+2x3 = 3 + 2x. See how we're isolating x? We’re just a few steps away!

Solving for the Final Value of x

We're in the home stretch, guys! Our equation is currently 3=3+2x3 = 3 + 2x. To get 2x2x by itself, we need to get rid of that constant term '+3' on the right side. We do this by subtracting 33 from both sides of the equation. So, we have 3βˆ’3=(3+2x)βˆ’33 - 3 = (3 + 2x) - 3. On the left side, 3βˆ’33 - 3 equals 00. On the right side, the +3+3 and βˆ’3-3 cancel each other out, leaving us with just 2x2x. So, our equation simplifies to 0=2x0 = 2x. The final step to solve for the value of x is to divide both sides by 22. This gives us 0/2=2x/20 / 2 = 2x / 2. And what do we get? 0=x0 = x. So, the value of xx is 00. We’ve successfully solved for x!

Checking Our Answer

It's always a smart move to double-check our answer to make sure we didn't mess up anywhere, right? Our solution is x=0x=0. Let's plug this value back into the original equation: 2(xβˆ’3)+9=3(x+1)+x2(x-3)+9=3(x+1)+x.

Left Side: Substitute x=0x=0 into 2(xβˆ’3)+92(x-3)+9. 2(0βˆ’3)+9=2(βˆ’3)+9=βˆ’6+9=32(0-3)+9 = 2(-3)+9 = -6+9 = 3.

Right Side: Substitute x=0x=0 into 3(x+1)+x3(x+1)+x. 3(0+1)+0=3(1)+0=3+0=33(0+1)+0 = 3(1)+0 = 3+0 = 3.

Since the left side (33) equals the right side (33), our answer x=0x=0 is correct! This verification step is super important, especially in exams, to ensure you've got the right answer. It really boosts your confidence when you see both sides match up perfectly.

Conclusion

So there you have it, guys! By breaking down the equation 2(xβˆ’3)+9=3(x+1)+x2(x-3)+9=3(x+1)+x into manageable steps – simplifying both sides, isolating the variable xx, and then solving for its value – we found that x=0x=0. This process highlights the importance of careful calculation and the distributive property in algebra. Remember, with practice, solving for x and other algebraic equations becomes second nature. Keep practicing, and you'll be a math whiz in no time!