Problem 69 - Elementary Algebra

Question

Solve the fractional equation \(\frac{x-1}{x+4}=\frac{x+3}{x-1}\).

Step-by-Step Solution

Answer

Question: Solve the fractional equation and check the solution: \(\frac{x-1}{x+4} = \frac{x+3}{x-1}\) Answer: x = -\(\frac{11}{9}\)

Step 1: Clear the denominators

Multiply both sides of the equation by \((x+4)(x-1)\) to eliminate the denominators: \((x+4)(x-1) \cdot \frac{x-1}{x+4} = (x+4)(x-1) \cdot \frac{x+3}{x-1}\)

Step 2: Simplify and collect like terms

Simplify the equation and cancel any common factors: \((x-1)^2 = (x+4)(x+3)\) Expand each side of the equation: \(x^2 - 2x + 1 = x^2 + 7x + 12\)

Step 3: Solve the equation

Subtract \(x^2\) from both sides and simplify the equation: \(-2x + 1 = 7x + 12\) Add 2x to both sides to isolate the x-term: \(1 = 9x + 12\) Subtract 12 from both sides to isolate the x-term: \(-11 = 9x\) Divide by 9 to solve for x: \(x = -\frac{11}{9}\)

Step 4: Check the solution

Check that the solution does not create any division by zero in the original equation: \(\frac{(-\frac{11}{9})-1}{(-\frac{11}{9})+4} \stackrel{?}{=} \frac{(-\frac{11}{9})+3}{(-\frac{11}{9})-1}\) Since neither the denominators are zero, we can proceed to check if the equation holds true: \(\frac{-\frac{20}{9}}{\frac{25}{9}} = \frac{\frac{16}{9}}{-\frac{20}{9}}\) Both fractions are equivalent, so the solution \(x=-\frac{11}{9}\) is valid.

Key Concepts

Clearing Denominators

One of the fundamental steps in solving fractional equations is to clear the denominators. The process involves finding a common multiple of the denominators and using it to eliminate the fractions. This simplifies the equation significantly.
To clear the denominators in the equation \( \frac{x-1}{x+4} = \frac{x+3}{x-1} \), we multiply both sides by the product of the denominators \((x+4)(x-1)\).
The multiplication cancels out the denominators, transforming the fractional equation into an equation with no fractions. This makes it much simpler to handle, as it becomes a standard algebraic equation. Once the denominators are cleared, you are left with expressions that can be further simplified and solved.

Expanding Equations

After clearing the denominators, the next step is expanding the equations. Expansion involves multiplying out the factors to simplify the equation.
From the cleared equation \((x-1)^2 = (x+4)(x+3)\), you need to expand both sides. For example:

• Expand \((x-1)^2\) to get \(x^2 - 2x + 1\).
• Expand \((x+4)(x+3)\) to get \(x^2 + 7x + 12\).

Expansion helps in writing the equation in a standard polynomial form, making it easier to solve for the unknown variable. Remember, careful expansion is crucial as any mistake could lead to wrong results.

Checking Solutions

Checking solutions is a necessary step to ensure that the solution found does not cause any invalid mathematical operations, like division by zero.
In our exercise, we find \(x = -\frac{11}{9}\) as a potential solution. It's important to replace \(x\) in the original equation to confirm the solution is valid:

• Substitute \(-\frac{11}{9}\) back into \(\frac{x-1}{x+4} = \frac{x+3}{x-1}\).
• Ensure that none of the denominators becomes zero due to this substitution, which would make the solution invalid.

If the equation holds true with the found value, and no zero denominator arises, the solution is correct and valid.

Simplifying Equations

After expanding, it is often necessary to simplify the equation to further isolate the variable. This may involve additional steps:

• Combine like terms on both sides.
• Move all terms to one side of the equation.

In our example, subtract \(x^2\) on both sides to simplify to \(-2x + 1 = 7x + 12\). From here, rearrange to isolate terms involving \(x\), using operations like addition or subtraction.
Finally, solve for \(x\) by dividing the isolated term by its coefficient. Always remember: simplifying helps in breaking down the equation to its most elementary form, making it straightforward to find the solution.