Simplify The Equation And Solve For U U U :${ -2(8u - 7) + 2u = 9 + 2(2u - 5) }$Find The Value Of U U U .

by ADMIN 107 views

Introduction

Solving algebraic equations is a fundamental concept in mathematics, and it requires a step-by-step approach to simplify the equation and isolate the variable. In this article, we will focus on simplifying the given equation and solving for the value of uu. We will use basic algebraic operations such as distributing, combining like terms, and isolating the variable to find the solution.

The Given Equation

The given equation is:

βˆ’2(8uβˆ’7)+2u=9+2(2uβˆ’5){ -2(8u - 7) + 2u = 9 + 2(2u - 5) }

Step 1: Distribute the Negative 2

To simplify the equation, we will start by distributing the negative 2 to the terms inside the parentheses.

βˆ’2(8uβˆ’7)=βˆ’16u+14{ -2(8u - 7) = -16u + 14 }

So, the equation becomes:

βˆ’16u+14+2u=9+2(2uβˆ’5){ -16u + 14 + 2u = 9 + 2(2u - 5) }

Step 2: Distribute the 2

Next, we will distribute the 2 to the terms inside the parentheses on the right-hand side of the equation.

2(2uβˆ’5)=4uβˆ’10{ 2(2u - 5) = 4u - 10 }

So, the equation becomes:

βˆ’16u+14+2u=9+4uβˆ’10{ -16u + 14 + 2u = 9 + 4u - 10 }

Step 3: Combine Like Terms

Now, we will combine like terms on both sides of the equation.

On the left-hand side, we have:

βˆ’16u+2u=βˆ’14u{ -16u + 2u = -14u }

And on the right-hand side, we have:

9βˆ’10=βˆ’1{ 9 - 10 = -1 }

So, the equation becomes:

βˆ’14u+14=βˆ’1+4u{ -14u + 14 = -1 + 4u }

Step 4: Isolate the Variable

To isolate the variable uu, we will move all the terms containing uu to one side of the equation and the constant terms to the other side.

First, we will add 14 to both sides of the equation to get rid of the constant term on the left-hand side.

βˆ’14u=βˆ’1+4uβˆ’14{ -14u = -1 + 4u - 14 }

Next, we will simplify the right-hand side of the equation.

βˆ’1βˆ’14=βˆ’15{ -1 - 14 = -15 }

And we will add 14 to the right-hand side of the equation.

βˆ’1+4uβˆ’14+14=βˆ’15+14{ -1 + 4u - 14 + 14 = -15 + 14 }

So, the equation becomes:

βˆ’14u=βˆ’1+4u{ -14u = -1 + 4u }

Step 5: Move the Variable Terms

Now, we will move the variable terms to the left-hand side of the equation and the constant terms to the right-hand side.

We will subtract 4u from both sides of the equation to get rid of the variable term on the right-hand side.

βˆ’14uβˆ’4u=βˆ’1{ -14u - 4u = -1 }

Next, we will simplify the left-hand side of the equation.

βˆ’18u=βˆ’1{ -18u = -1 }

Step 6: Solve for uu

Finally, we will solve for the value of uu by dividing both sides of the equation by -18.

u=βˆ’1βˆ’18{ u = \frac{-1}{-18} }

u=118{ u = \frac{1}{18} }

The final answer is 118\boxed{\frac{1}{18}}.

Conclusion

In this article, we simplified the given equation and solved for the value of uu. We used basic algebraic operations such as distributing, combining like terms, and isolating the variable to find the solution. The final answer is 118\boxed{\frac{1}{18}}.

Introduction

In our previous article, we simplified the given equation and solved for the value of uu. We received many questions from readers who were struggling to understand the steps involved in solving the equation. In this article, we will address some of the most frequently asked questions and provide additional explanations to help readers better understand the concept.

Q: What is the first step in solving the equation?

A: The first step in solving the equation is to distribute the negative 2 to the terms inside the parentheses. This will help us simplify the equation and make it easier to work with.

Q: Why do we need to combine like terms?

A: Combining like terms is an important step in solving the equation. It helps us simplify the equation and make it easier to isolate the variable. By combining like terms, we can eliminate unnecessary variables and make the equation more manageable.

Q: How do we isolate the variable?

A: To isolate the variable, we need to move all the terms containing the variable to one side of the equation and the constant terms to the other side. We can do this by adding or subtracting the same value to both sides of the equation.

Q: What is the final answer?

A: The final answer is 118\boxed{\frac{1}{18}}. This is the value of uu that satisfies the equation.

Q: Can you explain the concept of distributing in more detail?

A: Distributing is a basic algebraic operation that involves multiplying a value to each term inside a set of parentheses. For example, if we have the expression βˆ’2(8uβˆ’7)-2(8u - 7), we need to multiply the negative 2 to each term inside the parentheses, which gives us βˆ’16u+14-16u + 14.

Q: Can you explain the concept of combining like terms in more detail?

A: Combining like terms involves adding or subtracting terms that have the same variable and coefficient. For example, if we have the expression βˆ’16u+2u-16u + 2u, we can combine the two terms by adding their coefficients, which gives us βˆ’14u-14u.

Q: How do we know which side of the equation to move the variable terms to?

A: We need to move the variable terms to the side of the equation that has the fewest number of variable terms. This will make it easier to isolate the variable and solve for its value.

Q: Can you provide more examples of solving equations?

A: Yes, we can provide more examples of solving equations. Here are a few examples:

Example 1

Solve for xx in the equation:

2x+5=11{ 2x + 5 = 11 }

Solution

To solve for xx, we need to isolate the variable by moving the constant term to the other side of the equation.

2x=11βˆ’5{ 2x = 11 - 5 }

2x=6{ 2x = 6 }

Next, we need to divide both sides of the equation by 2 to solve for xx.

x=62{ x = \frac{6}{2} }

x=3{ x = 3 }

Example 2

Solve for yy in the equation:

3yβˆ’2=7{ 3y - 2 = 7 }

Solution

To solve for yy, we need to isolate the variable by moving the constant term to the other side of the equation.

3y=7+2{ 3y = 7 + 2 }

3y=9{ 3y = 9 }

Next, we need to divide both sides of the equation by 3 to solve for yy.

y=93{ y = \frac{9}{3} }

y=3{ y = 3 }

Conclusion

In this article, we addressed some of the most frequently asked questions and provided additional explanations to help readers better understand the concept of solving equations. We also provided more examples of solving equations to help readers practice and reinforce their understanding of the concept.