91Ó°ÊÓ

Algebraic manipulation is a fundamental skill in solving equations. It involves rearranging and simplifying expressions to isolate the variable you are solving for. In our problem, we start with the equation:






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Step 2 involves multiplying every term by x to eliminate the fraction: \br>


(x-\frac{12}{x}=4 )
Step 3: Move all terms to one side to set the equation to zero:


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EXERCISE: Solve. If no solution exists, state this. $$x-\frac{12}{x}=4$$ bru can simplify it: it's all the same.
Tou should always substitute the solutions back into the original equation to verify they are correct.

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-x - \frac{12}{-x}=4;- ex rx \frac{12}, it's basically so correct? It all comes together when you bring the algebra into play:

• Multiplying through by x eliminates the fraction, simplifying the equation.

Factoring polynomials is essential in solving quadratic equations. After rearranging our equation to




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( x^2 - 4x - 12 = 0 ), the next step is to factor it. This means finding pairs of numbers that multiply to give the constant term (-12) and add up to give the coefficient of the middle term (-4).

To successfully factor, consider what ( x - 6 ) and ( x + 2 ) equate to respectively.

• Factor:
  
  \( x\) - 6 and \( x\) + 2 respectively.

Now solve each factor by setting them to zero:

Verifying solutions is the final step to ensure the solutions found are correct. This involves substituting the solutions back into the original equation: Here, let's verify \(x=6\) and \(x=-2\).

-2 - \frac{12}{-2} = 4
When you substitute \(x=6\), you get \(6-\frac{12}{6}= 4\)
which is true. Similarly, substitute \(x=-2\), which also simplifies correctly to -2 leading to verification of both solutions:

Substitute \(x=6\)

substituting correctly \(x=2\) True.

Remember to always verify solutions are true: