Solve The Equation. 5 ( 5 + C ) = 4 ( 4 + C 5(5+c)=4(4+c 5 ( 5 + C ) = 4 ( 4 + C ] C = C = C =

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Introduction

In mathematics, solving equations is a fundamental concept that helps us find the value of unknown variables. Equations are statements that express the equality of two mathematical expressions. In this article, we will focus on solving a linear equation involving a variable $c$. The given equation is $5(5+c)=4(4+c)$, and we need to find the value of $c$.

Understanding the Equation

The given equation is a linear equation, which means it can be solved using basic algebraic operations. The equation is $5(5+c)=4(4+c)$. To solve this equation, we need to isolate the variable $c$ on one side of the equation.

Distributive Property

To start solving the equation, we can use the distributive property, which states that for any numbers $a$, $b$, and $c$, $a(b+c)=ab+ac$. We can apply this property to both sides of the equation.

5(5+c) = 5(5) + 5c
4(4+c) = 4(4) + 4c

Simplifying the Equation

Now that we have applied the distributive property, we can simplify the equation by combining like terms.

25 + 5c = 16 + 4c

Isolating the Variable

To isolate the variable $c$, we need to get all the terms involving $c$ on one side of the equation. We can do this by subtracting $4c$ from both sides of the equation.

25 + 5c - 4c = 16

Combining Like Terms

Now that we have subtracted $4c$ from both sides of the equation, we can combine like terms.

25 + c = 16

Solving for $c$

To solve for $c$, we need to isolate the variable $c$ on one side of the equation. We can do this by subtracting $25$ from both sides of the equation.

c = 16 - 25

Evaluating the Expression

Now that we have the expression $c = 16 - 25$, we can evaluate it to find the value of $c$.

c = -9

Conclusion

In this article, we solved the equation $5(5+c)=4(4+c)$ to find the value of the variable $c$. We used the distributive property to simplify the equation and then isolated the variable $c$ by combining like terms and subtracting $25$ from both sides of the equation. The final value of $c$ is $-9$.

Frequently Asked Questions

• What is the distributive property? The distributive property is a mathematical concept that states that for any numbers $a$, $b$, and $c$, $a(b+c)=ab+ac$.
• How do I solve a linear equation? To solve a linear equation, you need to isolate the variable on one side of the equation by combining like terms and using basic algebraic operations.
• What is the value of $c$ in the equation $5(5+c)=4(4+c)$? The value of $c$ in the equation $5(5+c)=4(4+c)$ is $-9$.

Related Topics

• Solving quadratic equations
• Linear equations
• Algebraic operations
• Distributive property

References

• [1] "Algebra" by Michael Artin
• [2] "Linear Algebra" by Jim Hefferon
• [3] "Mathematics for Computer Science" by Eric Lehman and Tom Leighton
  

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Introduction

In our previous article, we solved the equation $5(5+c)=4(4+c)$ to find the value of the variable $c$. In this article, we will answer some frequently asked questions related to solving the equation.

Q&A

Q: What is the distributive property?

A: The distributive property is a mathematical concept that states that for any numbers $a$, $b$, and $c$, $a(b+c)=ab+ac$. This property is used to simplify equations by distributing the terms inside the parentheses.

Q: How do I solve a linear equation?

A: To solve a linear equation, you need to isolate the variable on one side of the equation by combining like terms and using basic algebraic operations. You can use the distributive property to simplify the equation and then isolate the variable.

Q: What is the value of $c$ in the equation $5(5+c)=4(4+c)$?

A: The value of $c$ in the equation $5(5+c)=4(4+c)$ is $-9$. This was found by applying the distributive property, simplifying the equation, and isolating the variable $c$.

Q: Can I use the distributive property to solve any equation?

A: No, the distributive property can only be used to solve equations that involve parentheses. If the equation does not involve parentheses, you cannot use the distributive property to solve it.

Q: How do I know if an equation is linear or not?

A: An equation is linear if it can be written in the form $ax + b = c$, where $a$, $b$, and $c$ are constants and $x$ is the variable. If the equation cannot be written in this form, it is not linear.

Q: Can I use algebraic operations to solve any equation?

A: Yes, algebraic operations such as addition, subtraction, multiplication, and division can be used to solve any equation. However, you need to be careful when using these operations to ensure that you are not introducing extraneous solutions.

Q: What is an extraneous solution?

A: An extraneous solution is a solution to an equation that is not actually a solution. This can happen when you use algebraic operations to solve an equation, but the operations introduce a solution that is not actually valid.

Conclusion

In this article, we answered some frequently asked questions related to solving the equation $5(5+c)=4(4+c)$. We covered topics such as the distributive property, solving linear equations, and algebraic operations. We also discussed extraneous solutions and how to avoid them.

Frequently Asked Questions (FAQs)

• What is the distributive property?
• How do I solve a linear equation?
• What is the value of $c$ in the equation $5(5+c)=4(4+c)$?
• Can I use the distributive property to solve any equation?
• How do I know if an equation is linear or not?
• Can I use algebraic operations to solve any equation?
• What is an extraneous solution?

Related Topics

• Solving quadratic equations
• Linear equations
• Algebraic operations
• Distributive property
• Exponential equations

References

• [1] "Algebra" by Michael Artin
• [2] "Linear Algebra" by Jim Hefferon
• [3] "Mathematics for Computer Science" by Eric Lehman and Tom Leighton

Further Reading

• "Solving Equations" by Khan Academy
• "Linear Equations" by Mathway
• "Algebraic Operations" by Wolfram Alpha