Solving $-\frac{1121}{36}+\frac{94}{9} X=\frac{67}{12}(x-\frac{1}{2})+\frac{1}{2} X$

by Rajiv Sharma 85 views

Hey guys! Let's dive into solving a linear equation today. We've got a fun one here: βˆ’112136+949x=6712(xβˆ’12)+12x-\frac{1121}{36}+\frac{94}{9} x=\frac{67}{12}\left(x-\frac{1}{2}\right)+\frac{1}{2} x. Don't worry, it looks scarier than it is. We'll break it down step-by-step so it's super easy to follow. Grab your pencils, and let's get started!

1. Clearing Fractions: The Key to Simplicity

Okay, the first thing we see are fractions, and nobody likes dealing with those more than we have to. So, let’s get rid of them right away! The trick here is to find the least common multiple (LCM) of all the denominators. In our equation, the denominators are 36, 9, 12, and 2. What's the LCM of these numbers? It's 36. This means we're going to multiply both sides of the equation by 36. This will clear out all those pesky fractions and make our lives much easier. Remember, whatever we do to one side of the equation, we have to do to the other to keep it balanced. Think of it like a scaleβ€”if you add weight to one side, you have to add the same weight to the other to keep it level.

So, we multiply both sides of the equation βˆ’112136+949x=6712(xβˆ’12)+12x-\frac{1121}{36}+\frac{94}{9} x=\frac{67}{12}\left(x-\frac{1}{2}\right)+\frac{1}{2} x by 36:

36(βˆ’112136+949x)=36(6712(xβˆ’12)+12x)36\left(-\frac{1121}{36}+\frac{94}{9} x\right) = 36\left(\frac{67}{12}\left(x-\frac{1}{2}\right)+\frac{1}{2} x\right)

Now, we distribute the 36 on both sides. This means we multiply each term inside the parentheses by 36. On the left side, we have:

36βˆ—βˆ’112136+36βˆ—949x36 * -\frac{1121}{36} + 36 * \frac{94}{9}x

The 36s in the first term cancel out, leaving us with -1121. For the second term, 36 divided by 9 is 4, so we have 4 * 94x, which equals 376x. So, the left side simplifies to:

βˆ’1121+376x-1121 + 376x

On the right side, we have:

36βˆ—6712(xβˆ’12)+36βˆ—12x36 * \frac{67}{12}\left(x-\frac{1}{2}\right) + 36 * \frac{1}{2}x

For the first term, 36 divided by 12 is 3, so we have 3 * 67(x - \frac{1}{2}). For the second term, 36 divided by 2 is 18, so we have 18x. So the right side looks like this:

3βˆ—67(xβˆ’12)+18x3 * 67\left(x-\frac{1}{2}\right) + 18x

Let's simplify this further. 3 * 67 is 201, so we have:

201(xβˆ’12)+18x201\left(x-\frac{1}{2}\right) + 18x

Now, we distribute the 201 inside the parentheses:

201xβˆ’201βˆ—12+18x201x - 201 * \frac{1}{2} + 18x

Which simplifies to:

201xβˆ’2012+18x201x - \frac{201}{2} + 18x

So, our equation now looks like this:

βˆ’1121+376x=201xβˆ’2012+18x-1121 + 376x = 201x - \frac{201}{2} + 18x

See? Way less scary already! We've gotten rid of the fractions and simplified both sides. Now we are ready to collect the like terms.

2. Gathering Like Terms: Organizing the Equation

Now that we've cleared the fractions, our equation looks much cleaner: βˆ’1121+376x=201xβˆ’2012+18x-1121 + 376x = 201x - \frac{201}{2} + 18x. The next step is to gather the like terms. This means we want to bring all the terms with 'x' to one side of the equation and all the constant terms (the numbers) to the other side. It's like sorting your laundry – you put all the shirts together, all the pants together, and so on.

First, let's combine the 'x' terms on the right side of the equation. We have 201x and 18x. Adding those together gives us 219x. So, the equation becomes:

βˆ’1121+376x=219xβˆ’2012-1121 + 376x = 219x - \frac{201}{2}

Now, we want to get all the 'x' terms on one side. Let's subtract 219x from both sides of the equation. This will move the 'x' terms from the right side to the left side:

βˆ’1121+376xβˆ’219x=219xβˆ’2012βˆ’219x-1121 + 376x - 219x = 219x - \frac{201}{2} - 219x

On the right side, 219x - 219x cancels out, leaving us with just -\frac{201}{2}. On the left side, we combine 376x and -219x, which gives us 157x. So, our equation now looks like this:

βˆ’1121+157x=βˆ’2012-1121 + 157x = -\frac{201}{2}

Next, we want to get all the constant terms on the other side of the equation. To do this, we'll add 1121 to both sides:

βˆ’1121+157x+1121=βˆ’2012+1121-1121 + 157x + 1121 = -\frac{201}{2} + 1121

On the left side, -1121 + 1121 cancels out, leaving us with just 157x. On the right side, we need to add -\frac201}{2} and 1121. To do this, we need a common denominator. We can rewrite 1121 as a fraction with a denominator of 2 1121 = \frac{2242{2}. So, we have:

βˆ’2012+22422-\frac{201}{2} + \frac{2242}{2}

Adding these fractions gives us \frac{2041}{2}. So, our equation now looks like this:

157x=20412157x = \frac{2041}{2}

We're almost there! We've successfully gathered like terms, and now we have a much simpler equation to solve.

3. Isolating the Variable: Finding the Value of x

Okay, we're in the home stretch now! We've simplified the equation to 157x=20412157x = \frac{2041}{2}. Our goal now is to isolate 'x'. This means we want to get 'x' by itself on one side of the equation. To do this, we need to get rid of the 157 that's multiplying 'x'.

The way we do that is by dividing both sides of the equation by 157. Remember, whatever we do to one side, we have to do to the other to keep the equation balanced:

157x157=20412157\frac{157x}{157} = \frac{\frac{2041}{2}}{157}

On the left side, 157 divided by 157 is 1, so we're left with just 'x'. On the right side, we have a fraction divided by a whole number. To divide a fraction by a whole number, we can multiply the denominator of the fraction by the whole number. So, we have:

x=20412βˆ—157x = \frac{2041}{2 * 157}

Now, we multiply 2 and 157 in the denominator, which gives us 314. So, we have:

x=2041314x = \frac{2041}{314}

Now, we need to simplify this fraction. We can see if 2041 is divisible by 157. If you divide 2041 by 157, you get 13. So, we can simplify the fraction:

x=157βˆ—132βˆ—157x = \frac{157 * 13}{2 * 157}

The 157s cancel out, leaving us with:

x=132x = \frac{13}{2}

And there we have it! We've isolated 'x', and we found that x = \frac{13}{2}. That's our solution!

4. Verification: Ensuring the Accuracy

Before we celebrate, let's do a quick check to make sure our answer is correct. This is a crucial step in solving any equation. We're going to plug our solution, x=132x = \frac{13}{2}, back into the original equation and see if both sides are equal. Remember our original equation:

βˆ’112136+949x=6712(xβˆ’12)+12x-\frac{1121}{36}+\frac{94}{9} x=\frac{67}{12}\left(x-\frac{1}{2}\right)+\frac{1}{2} x

Now, let's substitute x=132x = \frac{13}{2} into the equation:

βˆ’112136+949βˆ—132=6712(132βˆ’12)+12βˆ—132-\frac{1121}{36}+\frac{94}{9} * \frac{13}{2}=\frac{67}{12}\left(\frac{13}{2}-\frac{1}{2}\right)+\frac{1}{2} * \frac{13}{2}

First, let's simplify the left side:

βˆ’112136+949βˆ—132=βˆ’112136+94βˆ—139βˆ—2=βˆ’112136+122218-\frac{1121}{36}+\frac{94}{9} * \frac{13}{2} = -\frac{1121}{36} + \frac{94 * 13}{9 * 2} = -\frac{1121}{36} + \frac{1222}{18}

To add these fractions, we need a common denominator. The LCM of 36 and 18 is 36. So, we rewrite \frac{1222}{18} with a denominator of 36:

122218=1222βˆ—218βˆ—2=244436\frac{1222}{18} = \frac{1222 * 2}{18 * 2} = \frac{2444}{36}

Now we can add the fractions:

βˆ’112136+244436=2444βˆ’112136=132336-\frac{1121}{36} + \frac{2444}{36} = \frac{2444 - 1121}{36} = \frac{1323}{36}

So, the left side simplifies to \frac{1323}{36}.

Now, let's simplify the right side:

6712(132βˆ’12)+12βˆ—132=6712(13βˆ’12)+134=6712βˆ—122+134\frac{67}{12}\left(\frac{13}{2}-\frac{1}{2}\right)+\frac{1}{2} * \frac{13}{2} = \frac{67}{12}\left(\frac{13-1}{2}\right)+\frac{13}{4} = \frac{67}{12} * \frac{12}{2} + \frac{13}{4}

6712βˆ—122=67βˆ—1212βˆ—2=672\frac{67}{12} * \frac{12}{2} = \frac{67 * 12}{12 * 2} = \frac{67}{2}

So, we have:

672+134\frac{67}{2} + \frac{13}{4}

To add these fractions, we need a common denominator. The LCM of 2 and 4 is 4. So, we rewrite \frac{67}{2} with a denominator of 4:

672=67βˆ—22βˆ—2=1344\frac{67}{2} = \frac{67 * 2}{2 * 2} = \frac{134}{4}

Now we can add the fractions:

1344+134=134+134=1474\frac{134}{4} + \frac{13}{4} = \frac{134 + 13}{4} = \frac{147}{4}

So, the right side simplifies to \frac{147}{4}.

Now, let's see if both sides are equal:

Left side: \frac{1323}{36}

Right side: \frac{147}{4}

To compare these, we can simplify \frac{1323}{36} by dividing both numerator and denominator by 9 which gives us \frac{147}{4}

So, both sides are equal! This means our solution, x=132x = \frac{13}{2}, is correct.

Conclusion: Mastering Linear Equations

Awesome job, guys! We've successfully solved the linear equation βˆ’112136+949x=6712(xβˆ’12)+12x-\frac{1121}{36}+\frac{94}{9} x=\frac{67}{12}\left(x-\frac{1}{2}\right)+\frac{1}{2} x. We tackled it step-by-step, from clearing fractions to isolating the variable and verifying our solution. Remember, the key to solving linear equations is to stay organized, follow the steps carefully, and always double-check your work.

Solving equations like this might seem intimidating at first, but with practice, you'll become a pro. You've got this! Keep practicing, and you'll be able to solve any linear equation that comes your way. And remember, math can be fun when you break it down into manageable steps. Keep up the great work!