15 Aug, 2024
· Mathematics

Which equation is the inverse of (x-4)^2-2/3=6y-12

Short Answer
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Long Explanation

Explanation

To find the inverse of the given equation:

(x4)223=6y12(x - 4)^2 - \frac{2}{3} = 6y - 12

Step 1: Simplify the Original Equation

First, we start by isolating yy on one side of the equation. Add 1212 to both sides:

(x4)223+12=6y(x - 4)^2 - \frac{2}{3} + 12 = 6y

Simplify:

(x4)2+1113=6y(x - 4)^2 + 11\frac{1}{3} = 6y

Note that:

1113=34311\frac{1}{3} = \frac{34}{3}

Thus:

(x4)2+343=6y(x - 4)^2 + \frac{34}{3} = 6y

Multiply through by 3:

3(x4)2+34=18y3(x - 4)^2 + 34 = 18y

Divide both sides by 18:

y=118[3(x4)2+34]y = \frac{1}{18}[3(x - 4)^2 + 34]

Step 2: Solve for xx in Terms of yy

Swap xx and yy:

x=118[3(y4)2+34]x = \frac{1}{18}[3(y - 4)^2 + 34]

Multiply by 18:

18x=3(y4)2+3418x = 3(y - 4)^2 + 34

Subtract 34 from both sides:

18x34=3(y4)218x - 34 = 3(y - 4)^2

Divide by 3:

6x343=(y4)26x - \frac{34}{3} = (y - 4)^2

Step 3: Rewrite the Inverse Equation

Finally, solve for yy:

(y4)2=6x343(y - 4)^2 = 6x - \frac{34}{3}

Take the square root of both sides:

y4=±6x343y - 4 = \pm \sqrt{6x - \frac{34}{3}}

Add 4 to both sides:

y=4±6x343y = 4 \pm \sqrt{6x - \frac{34}{3}}

Thus, the inverse function is:

y=4±6x343\boxed{y = 4 \pm \sqrt{6x - \frac{34}{3}}}

This gives us both possible values for the inverse function, depending on whether the positive or negative square root is taken.

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Emily Rosen

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Emily Rosen is a recent graduate with a Master's in Mathematics from the University of Otago. She has been tutoring math students and working as a part-time contract writer for the past three years. She is passionate about helping students overcome their fear of math through easily digestible content.

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Concept

Key Concepts Used: - Solving And Simplifying Algebraic Equations

Understanding Key Concepts Used in Solving and Simplifying Algebraic Equations

In algebra, solving and simplifying equations are essential skills. These processes help you find the values of unknown variables and streamline complex expressions. Below, we explore some of the key concepts that are vital in this context.

Key Concepts

  1. Variables and Constants

    • Variables: Symbols (like xx or yy) representing unknown values.
    • Constants: Fixed numerical values.
  2. Operations

    • Addition, subtraction, multiplication, and division.
    • Use of parentheses ((( and ))) for grouping terms.
  3. Balancing Equations

    • Ensuring both sides of the equation are equal.
    • Whatever operation you perform on one side, must be performed on the other side too.

Techniques

  1. Isolating Variables
    • Move terms involving variables to one side, constants to the other.
    • For example:
3x+5=203x=2053x=153x + 5 = 20 \rightarrow 3x = 20 - 5 \rightarrow 3x = 15 \rightarrow x=153x=5\rightarrow x = \frac{15}{3} \rightarrow x = 5
  1. Combining Like Terms

    • Terms with the same variable and power can be combined.
    • Example: 2x+3x=5x2x + 3x = 5x
  2. Factoring

    • Factoring polynomial expressions to simplify and solve.
    • Quadratic example: x2+5x+6=0(x+2)(x+3)=0x^2 + 5x + 6 = 0 \rightarrow (x + 2)(x + 3) = 0
    • Solving: x+2=0x=2x + 2 = 0 \rightarrow x = -2 x+3=0x=3x + 3 = 0 \rightarrow x = -3
  3. Using the Distributive Property

    • Distributing a factor across terms inside parentheses.
    • Example: a(b+c)=ab+aca(b + c) = ab + ac
    • Solving example:
2(x+3)=142x+6=142x=82(x + 3) = 14 \rightarrow 2x + 6 = 14 \rightarrow 2x = 8 \rightarrow x=82x=4\rightarrow x = \frac{8}{2} \rightarrow x = 4

Simplifying Expressions

  1. Canceling Common Factors

    • Simplify fractions by canceling common factors in the numerator and the denominator. 6x3=2x\frac{6x}{3} = 2x
  2. Combining Fractions

    • Combine fractions by finding a common denominator: 14+13=312+412=712\frac{1}{4} + \frac{1}{3} = \frac{3}{12} + \frac{4}{12} = \frac{7}{12}
  3. Rationalizing Denominators

    • Eliminate radicals from denominators: 12×22=22\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}

Quadratic Formula

For solving quadratic equations of the form

ax2+bx+c=0ax^2 + bx + c = 0 x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Conclusion

Mastering these concepts and techniques is crucial for solving and simplifying algebraic equations effectively. Practice regularly to enhance your algebraic problem-solving skills.

Concept

- Taking The Square Root Of Both Sides

Explanation of Taking the Square Root of Both Sides

Taking the square root of both sides of an equation is a common technique used to solve for a variable. This method is particularly useful when dealing with quadratic equations or when the variable is squared.

Basic Idea

If you have an equation where a variable is squared, such as:

x2=kx^2 = k

To solve for xx, you would take the square root of both sides:

x2=k\sqrt{x^2} = \sqrt{k}

Important Note on Solutions

It's crucial to remember that taking the square root of both sides introduces a ±\pm (plus or minus) symbol, because both the positive and negative values will satisfy the original squared equation. Hence, the actual equation becomes:

x=±kx = \pm\sqrt{k}

Example

Consider the equation:

x2=16x^2 = 16

By taking the square root of both sides, you get:

x=±16x = \pm\sqrt{16} x=±4x = \pm 4

Therefore, the solutions to the equation x2=16x^2 = 16 are x=4x = 4 and x=4x = -4.

Application Steps

  1. Isolate the squared term: Make sure the term containing the squared variable is by itself on one side of the equation.
  2. Take the square root of both sides: Apply the square root to both sides of the equation.
  3. Simplify and solve: Simplify the resulting expression and include both positive and negative solutions.

Constraints

When taking the square root of both sides, ensure that the value inside the square root (the radicand) is non-negative if dealing with real numbers. For complex solutions, different rules apply.

Conclusion

This process is a powerful and essential tool in algebra. By understanding and correctly applying the rule of taking the square root of both sides, you can effectively solve various types of equations, ensuring to not forget the ±\pm to capture all possible solutions.