Question

Solve the equation by cross multiplying. Check your solution(s). $$\frac{-1}{x-3}=\frac{x-4}{x^2-27}$$

Short answer

The valid solution for the equation is \(x = -5\). Other value \(x = 3\) is not a solution because it makes the original equation undefined.

Step-by-step solution

Cross-multiplication

Cross multiplication involves multiplying the numerator of one fraction by the denominator of the other fraction, and then setting that product equal to the product of the other numerator and denominator. So \((-1)*(x^2 - 27)\) is set equal to \((x - 4)*(x - 3)\) resulting in the equation \(-x^2 + 27 = x^2 - 7x + 12\).

Rearrange and combine similar terms

Rearrange the equation and combine like terms. This involves moving some terms around in order to get all the \(x\) terms together. This gives us: \(-x^2 -x^2 + 7x + 27 - 12 = 0\), which simplifies to \(-2x^2 + 7x + 15 = 0\).

Solving the Quadratic equation

The quadratic equation obtained in the previous step can be solved for \(x\). The Quadratic formula is the best method to solve this type of equation. Apply \(x = \frac{-b ± √(b^2 - 4ac)}{2a}\) where \(a=-2\), \(b=7\), and \(c=15\). This results in \(x = \frac{-7 ± √((7)^2 - 4*(-2)*15)}{2*(-2)}\), which gives \(x = 3\) or \(x = -5\)

Check the solution

Now we need to ensure that these solutions are not extraneous by substitifying both \(x = 3\) and \(x = -5\) into the original equation. \[-1/(3 - 3) = (3 - 4)/(3^2 - 27)\] is undefined, so \(x = 3\) isn't a solution. While \[-1/(-5 - 3) = (-5 - 4)/((-5)^2 - 27) = 1/8\] shows \(x = -5\) is a valid solution.

Key concepts

Cross Multiplication

When faced with an equation involving fractions, cross multiplication can be a handy tool. Essentially, it allows us to eliminate the fractions, turning a fraction equation into a more straightforward equation format.
Imagine two fractions set as equal, like \(-1/(x-3) = (x-4)/(x^2 - 27)\). Cross multiplication involves multiplying the numerator of each fraction with the denominator of the other. Thus, the operation would be:

• Numerator of the first fraction times the denominator of the second fraction: \(-1 \times (x^2 - 27)\)
• Numerator of the second fraction times the denominator of the first fraction: \((x-4) \times (x-3)\)

By cross multiplying, you convert the equation into: \(-x^2 + 27 = x^2 - 7x + 12\), which is much easier to work with.
Once this transformation occurs, you're free from dealing with pesky fractions and can continue solving for the variable.

Quadratic Equations

In our journey to solve the given equation, we ended up with \(-2x^2 + 7x + 15 = 0\). This is a classic form of a quadratic equation, which typically takes the form of \ (ax^2 + bx + c = 0)\.
Quadratic equations can often be solved using several methods—factoring, completing the square, or the quadratic formula. In this context, applying the quadratic formula is straightforward and reliable, especially when the equation doesn't easily factor.
The quadratic formula is given by:

• \x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\

Here, \(a = -2\), \(b = 7\), and \(c = 15\), and plugging these values into the formula helps solve for \(x\). You will find two potential solutions: \(x = 3\) and \(x = -5\).
It's crucial to understand that quadratic equations can provide two solutions because they represent parabolas, which can intersect the x-axis at two points.

Check Solutions

After obtaining potential solutions from a quadratic equation, you must verify if they genuinely satisfy the original equation. This process ensures that you haven't encountered any 'extraneous solutions'—solutions that emerge from the solving process but don’t solve the original equation.
To check the solutions, substitute each one back into the original equation \(-1/(x-3) = (x-4)/(x^2 - 27)\).

• For \(x = 3\), substitution results in division by zero, rendering this solution invalid.
• For \(x = -5\), the substitution yields a true statement, confirming it as a valid solution.

Checking solutions is a vital step to ensure accuracy and avoid errors, especially in complex algebraic equations. It allows us to conclude confidently about the validity of our found solutions.

Algebra 2

Algebra 2 expands upon the concepts learned in basic algebra, introducing more complex topics and problem-solving techniques. In this context, we've seen it applied through cross multiplication and solving quadratic equations.
Key concepts in Algebra 2 include:

• Manipulating and solving equations and inequalities
• Understanding functions and their behaviors
• Working with polynomials and rational expressions

Each concept builds upon a student's understanding, preparing them for higher-level mathematics by emphasizing problem-solving and logical reasoning.
In the exercise, we've integrated several core Algebra 2 skills—cross multiplying to simplify equations and employing the quadratic formula to find solutions. These are crucial skills that empower students to tackle more advanced mathematical challenges in the future.