Solve The Equation: $ Y + 6 = -3y + 26 }$Choose The Correct Value For { Y $}$ A. { Y = -8 $ $ B. { Y = -5 $}$ C. { Y = 5 $}$ D. { Y = 8 $}$

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Introduction

Linear equations are a fundamental concept in mathematics, and solving them is a crucial skill for students to master. In this article, we will focus on solving a specific linear equation, ${ y + 6 = -3y + 26 }$, and provide a step-by-step guide on how to find the correct value of yy.

Understanding Linear Equations

A linear equation is an equation in which the highest power of the variable(s) is 1. In other words, it is an equation that can be written in the form ax+b=cax + b = c, where aa, bb, and cc are constants, and xx is the variable. Linear equations can be solved using various methods, including algebraic manipulation, graphing, and substitution.

The Given Equation

The given equation is ${ y + 6 = -3y + 26 }$. This equation is a linear equation in one variable, yy. Our goal is to solve for yy.

Step 1: Isolate the Variable

To solve for yy, we need to isolate the variable on one side of the equation. We can do this by adding or subtracting the same value to both sides of the equation. In this case, we can add 3y3y to both sides of the equation to get:

y+6+3y=−3y+26+3y{ y + 6 + 3y = -3y + 26 + 3y }

Simplifying the equation, we get:

4y+6=26{ 4y + 6 = 26 }

Step 2: Subtract the Constant

Next, we need to subtract the constant term from both sides of the equation. In this case, we can subtract 6 from both sides of the equation to get:

4y+6−6=26−6{ 4y + 6 - 6 = 26 - 6 }

Simplifying the equation, we get:

4y=20{ 4y = 20 }

Step 3: Divide Both Sides

Finally, we need to divide both sides of the equation by the coefficient of the variable. In this case, we can divide both sides of the equation by 4 to get:

4y4=204{ \frac{4y}{4} = \frac{20}{4} }

Simplifying the equation, we get:

y=5{ y = 5 }

Conclusion

In this article, we solved the linear equation ${ y + 6 = -3y + 26 }$ using algebraic manipulation. We isolated the variable, subtracted the constant, and divided both sides of the equation to find the correct value of yy. The final answer is ${ y = 5 }$. This value is the correct solution to the given equation.

Choosing the Correct Answer

Now that we have solved the equation, we can choose the correct answer from the options provided:

A. ${ y = -8 }$ B. ${ y = -5 }$ C. ${ y = 5 }$ D. ${ y = 8 }$

The correct answer is C. ${ y = 5 }$. This is the value we obtained by solving the equation.

Practice Problems

Solving linear equations is a crucial skill that requires practice to master. Here are a few practice problems to help you improve your skills:

  1. Solve the equation ${ 2x + 3 = 5x - 2 }$.
  2. Solve the equation ${ x - 4 = 3x + 2 }$.
  3. Solve the equation ${ 4y - 2 = 2y + 10 }$.

Conclusion

Introduction

In our previous article, we solved a linear equation using algebraic manipulation. We isolated the variable, subtracted the constant, and divided both sides of the equation to find the correct value of yy. In this article, we will provide a Q&A guide to help you understand the concepts and techniques involved in solving linear equations.

Q: What is a linear equation?

A: A linear equation is an equation in which the highest power of the variable(s) is 1. In other words, it is an equation that can be written in the form ax+b=cax + b = c, where aa, bb, and cc are constants, and xx is the variable.

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. You can do this by adding or subtracting the same value to both sides of the equation, and then dividing both sides of the equation by the coefficient of the variable.

Q: What is the difference between a linear equation and a quadratic equation?

A: A linear equation is an equation in which the highest power of the variable(s) is 1, while a quadratic equation is an equation in which the highest power of the variable(s) is 2. For example, the equation x+2=5x + 2 = 5 is a linear equation, while the equation x2+2x+1=0x^2 + 2x + 1 = 0 is a quadratic equation.

Q: How do I know if an equation is linear or quadratic?

A: To determine if an equation is linear or quadratic, you need to look at the highest power of the variable(s) in the equation. If the highest power is 1, the equation is linear. If the highest power is 2, the equation is quadratic.

Q: What is the coefficient of a variable?

A: The coefficient of a variable is the number that is multiplied by the variable. For example, in the equation 2x+3=52x + 3 = 5, the coefficient of xx is 2.

Q: How do I isolate the variable in a linear equation?

A: To isolate the variable in a linear equation, you need to add or subtract the same value to both sides of the equation, and then divide both sides of the equation by the coefficient of the variable.

Q: What is the difference between adding and subtracting the same value to both sides of an equation?

A: Adding the same value to both sides of an equation is equivalent to subtracting the same value from both sides of the equation. For example, adding 2 to both sides of the equation x+3=5x + 3 = 5 is equivalent to subtracting 2 from both sides of the equation x+3=5x + 3 = 5.

Q: How do I divide both sides of an equation by the coefficient of the variable?

A: To divide both sides of an equation by the coefficient of the variable, you need to multiply both sides of the equation by the reciprocal of the coefficient. For example, if the coefficient of xx is 2, you need to multiply both sides of the equation by 12\frac{1}{2}.

Q: What is the reciprocal of a number?

A: The reciprocal of a number is 1 divided by the number. For example, the reciprocal of 2 is 12\frac{1}{2}.

Conclusion

In this article, we provided a Q&A guide to help you understand the concepts and techniques involved in solving linear equations. We covered topics such as the definition of a linear equation, how to solve a linear equation, and the difference between a linear equation and a quadratic equation. We also provided examples and explanations to help you understand the concepts.