Problem 64 - Algebra for College Students

Question

Solve each equation. $$\frac{1}{x-2}-\frac{2}{x+3}=\frac{11}{x^{2}+x-6}$$

Step-by-Step Solution

Answer

The solution is \(x = -4\).

Step 1: Identify the Common Denominator

First, notice that the left-hand side of the equation needs a common denominator. The denominators are \(x-2\) and \(x+3\). The right-hand side already has \(x^2 + x - 6\), which factors to \( (x-2)(x+3)\).

Step 2: Rewrite Each Fraction

Rewrite each fraction with the common denominator \ ((x-2)(x+3)) \. Multiply the numerator and denominator of each fraction on the left side by the missing factor: \[\frac{(x+3)}{(x-2)(x+3)} - \frac{2(x-2)}{(x-2)(x+3)} = \frac{11}{(x-2)(x+3)}\]

Step 3: Combine the Fractions

Combine the fractions on the left side, keeping the common denominator: \[\frac{x+3 - 2(x-2)}{(x-2)(x+3)} = \frac{11}{(x-2)(x+3)}\]Simplify the numerator: \[\frac{x+3 - 2x + 4}{(x-2)(x+3)}\]\[\frac{-x + 7}{(x-2)(x+3)} = \frac{11}{(x-2)(x+3)}\]

Step 4: Set the Numerators Equal

Since the denominators are the same on both sides, set the numerators equal to each other and solve for \(x\): \[-x + 7 = 11\]

Step 5: Solve for x

To isolate \(x\), subtract 7 from both sides: \[-x = 4\]Then multiply both sides by -1: \[x = -4\]

Step 6: Check the Solution

Check the solution by substituting \(x = -4\) back into the original equation to verify: \[\frac{1}{-4-2} - \frac{2}{-4+3} = \frac{11}{(-4)^2 + (-4) - 6}\]Simplify each term to check if both sides of the equation are equal.

Key Concepts

Common Denominator

When solving rational equations, finding a common denominator is key. It's similar to adding fractions, where you need to have the same bottom part of the fraction.
In our exercise, we have three denominators: \(x-2\), \(x+3\), and \(x^2 + x - 6\). The last denominator, \(x^2 + x - 6\), factors into \((x-2)(x+3)\).
This shows us that the common denominator for all the fractions is \((x-2)(x+3)\).
So, in the next step, we rewrite each fraction to have \((x-2)(x+3)\) as the denominator.
This makes it easier to combine and solve the equation.

Factoring

Factoring is like breaking down a number or expression into simpler, multiplied components.
For instance, the expression \(x^2 + x - 6\) is factored into \((x-2)(x+3)\).
Factoring helps us reveal common denominators, making the equation easier to solve.
You'll often need to factor quadratic expressions in rational equations to simplify them.
If you can't factor it easily by looking, remember the quadratic formula can help:

• Identify values of \(a\), \(b\), and \(c\) in your quadratic equation \(ax^2 + bx + c\).
• Then use the formula: \( \frac{-b \pm \sqrt{b^2-4ac}}{2a} \).

Once factored, you can rewrite the fractions with a common denominator, like we did in the solved exercise.

Setting Numerators Equal

After rewriting all fractions with a common denominator, you can set their numerators equal.
This step is crucial because if two fractions have the same denominator, their numerators must be equal for the fractions to be equal.
In our example, we have:

• Numerator on the left: \(-x + 7\)
• Numerator on the right: 11

So, we set \(-x + 7 = 11\).
Now solving for \(x\) is straightforward: subtract 7 from both sides, and then multiply by \(-1\).
This gives us the solution \(x = -4\).
Don't forget to check your solution by plugging it back into the original equation to ensure both sides are equal, confirming that your solution is correct!