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Algebra - "Solve for x" Challenge

Problem Statement

Solve for \(x\) and \(y\). You should have coordinate pairs (\(x\), \(y\)) for each equation:

\[ 2x^{4/3} + 5x^{3/4} = 6x^{4/3} - 3x^{3/4} \]

Step-by-Step Solution

To solve the equation:

\[ 2x^{4/3} + 5x^{3/4} = 6x^{4/3} - 3x^{3/4} \]

we can follow these steps:

Step 1: Combine like terms

First, let's group the terms involving \(x^{4/3}\) on one side and \(x^{3/4}\) on the other side. We do this by subtracting \(6x^{4/3}\) and adding \(3x^{3/4}\) to both sides of the equation:

\[ 2x^{4/3} + 5x^{3/4} - 6x^{4/3} + 3x^{3/4} = 0 \]

Combine like terms:

[ (2x^{4/3} - 6x^{4/3}) + (5x^{3/4} + 3x^{3/4}) = 0 ] [ -4x^{4/3} + 8x^{3/4} = 0 ]

Step 2: Isolate one of the terms

To simplify, isolate one of the terms. Here, isolate \(x^{4/3}\) or \(x^{3/4}\). Let's isolate \(x^{4/3}\):

\[ -4x^{4/3} = -8x^{3/4} \]

Divide both sides by -4:

\[ x^{4/3} = 2x^{3/4} \]

Step 3: Solve for \(x\)

To solve for \(x\), we need to eliminate the exponents. We can do this by dividing both sides by \(x^{3/4}\) to get:

\[ \frac{x^{4/3}}{x^{3/4}} = 2 \]

Apply the Quotient Rule for Exponents which states \(\frac{a^m}{a^n} = a^{m-n}\):

\[ x^{4/3 - 3/4} = 2 \]

First, find a common denominator to subtract the exponents:

\[ 4/3 - 3/4 \]

Convert to a common denominator (which is 12):

\[ \frac{16}{12} - \frac{9}{12} = \frac{7}{12} \]

So:

\[ x^{7/12} = 2 \]

Step 4: Solve for \(x\)

Raise both sides to the power of the reciprocal of \(7/12\) to solve for \(x\):

\[ x = 2^{12/7} \]

Step 5: Simplify if necessary

This is already in its simplest form.

Solution:

The solution to the equation is:

\[ x = 2^{12/7} \]

Summary of Properties Used:

  1. Combining Like Terms: To simplify the equation by grouping similar terms.
  2. Isolating Variables: To simplify the equation to a form where the variable can be solved.
  3. Quotient Rule for Exponents: To combine terms with exponents and solve for \(x\).
  4. Exponentiation: To solve for \(x\) by raising both sides to the power of the reciprocal.