Question

Solve the equation by cross multiplying. Check your solution(s). $$\frac{-2}{x-1}=\frac{x-8}{x+1}$$

Short answer

The solutions to the equation are \(x=2\) and \(x=5\).

Step-by-step solution

Cross Multiply

Cross multiply \(\frac{-2}{x-1}\) and \(\frac{x-8}{x+1}\) to get rid of the fractions. The cross multiplication results in \(-2 * (x+1) = (x-1) * (x-8)\). Simplify this to \(-2x -2 = x^2 - 9x + 8\).

Rearrange Equation

Rearrange the equation into a standard quadratic equation form. Transfer all terms to one side of the equation, resulting in \(x^2 - 7x +10 = 0\).

Factor the Quadratic

Factor the quadratic \(x^2 - 7x +10 = 0\) to \( (x-2)(x-5) = 0\).

Solve for X

Set each factor equal to zero and solve for x, giving solutions \(x=2\) and \(x=5\).

Check the Solutions

Insert these found x-values into the original equation to ensure that both are valid solutions. Any solution value which makes the denominator zero in the original equation is not valid. In this case, both possible solutions are valid.

Key concepts

Cross Multiplication

Cross multiplication is a method often used to solve equations that involve fractions. It helps eliminate the fractions by creating a direct equation between the two numerator and denominator pairs.

Here's how you apply it:

• Look at the equation \( \frac{-2}{x-1} = \frac{x-8}{x+1} \).
• The idea is to "cross" multiply: Multiply the numerator of each fraction by the denominator of the other fraction.

This results in two products that are set equal to each other:

• \(-2 \times (x+1)\)
• and \((x-8) \times (x-1)\)

Performing this operation will give you a new equation without fractions: \(-2(x+1) = (x-1)(x-8)\).

This simplifies the solving process and makes it easier to manipulate the equation further.

Keep in mind:

• Cross multiplication can only be used when you're dealing with a single equality between two fractions.
• It's crucial to ensure that both denominators in the fractions are not equal to zero to avoid undefined expressions.

Factoring Quadratic Equations

Once an equation is rearranged into a standard quadratic form, the next step usually is to factor it. A quadratic equation is generally written as \( ax^2 + bx + c = 0 \). Factoring is a way of breaking down the quadratic equation into simpler terms, making it easier to find solutions.

For instance, take the quadratic equation from our problem: \(x^2 - 7x + 10 = 0\).

To factor this, we look for two numbers that multiply to give the constant term (10 in this case) and add up to the linear coefficient (-7).

• The numbers that meet these criteria are -2 and -5.
• This means we can factor the quadratic as \((x-2)(x-5) = 0\).

Once factored, the equation becomes much simpler to solve. Each parenthetical expression is set to zero, leading to the solutions \( x = 2 \) and \( x = 5 \).
The beauty of this method lies in its ability to transform a seemingly complex equation into easy-to-solve factors.

Always check your solution for correctness after factoring to verify its validity.

Validity of Solutions

After solving an equation, it is important to check if the solutions you found are actually valid. This is a critical final step because certain values might not make sense within the given context of the problem.

For equations like the ones solved by cross multiplication, one of the main concerns is ensuring that none of the solutions make any denominator zero in the original equation.

• Returning to our original equation, \(\frac{-2}{x-1} = \frac{x-8}{x+1}\),\
• neither \(x-1\) nor \(x+1\) should be equal to zero.

This means that \(x eq 1\) and \(x eq -1\). If any solution equals these values, it must be discarded as it would lead to undefined expressions in the original fractions.

In our problem, substituting the solutions \( x = 2 \) and \( x = 5 \) back into the original equation confirms they are valid, since neither of them makes any denominator zero.
Checking your solutions against such constraints ensures a comprehensive understanding of the equation's behavior and prevents logical errors.