Problem 32 - Algebra 1: Concepts and Skills

Question

Solve the equation by multiplying each side by the least common denominator. Check your solutions. \(\frac{5}{3}+\frac{250}{9 r}=\frac{r}{9}\)

Step-by-Step Solution

Answer

The solutions to the equation, after checking, are \( r = 25 \) and \( r = -10 \)

Step 1: Identifying the Least Common Denominator (LCD)

The given equation is \( \frac{5}{3} + \frac{250}{9r} = \frac{r}{9} \). The denominators are \(3\), \(9r\), and \(9\). The least common denominator (LCD) among \(3\), \(9r\), and \(9\) would be \(9r\) as both 3 and 9 are factors of any multiple of \(9r\).

Step 2: Multiply Every Term by the LCD

Each term in the equation is multiplied by \(9r\), yielding: \( 9r * \frac{5}{3} + 9r * \frac{250}{9r} = 9r * \frac{r}{9} \), which simplifies to: \( 15r + 250 = r^2 \). After simplifying, The equation becomes a quadratic equation.

Step 3: Rearranging Equation

Rearrange the equation to standard form of a quadratic equation (i.e., \(0 = ax^2 + bx + c\)), which gives \( r^2 - 15r - 250 = 0 \).

Step 4: Solve for \(r\)

Solve the quadratic equation \( r^2 - 15r - 250 = 0 \) using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -15 \), and \( c = -250 \) to yield two possible solutions for \( r \).

Step 5: Check Solution

Substitute the solutions obtained from the previous step into the original equation to verify both solutions.

Key Concepts

Solving Rational Equations

Solving rational equations involves finding the value of the variable that makes the equation true. A rational equation is simply one that contains fractions with variables in their denominators, just like in our problem: \(\frac{5}{3} + \frac{250}{9r} = \frac{r}{9}\).

Here are the basic steps to solve rational equations:

• Identify the least common denominator (LCD) of all the fractions.
• Multiply every term by the LCD to eliminate the fractions.
• Simplify the resulting equation and solve for the variable.

The LCD in this context was found to be \(9r\). Multiplying every term by this LCD allows the fractions to be removed, thus making it easier to manipulate the equation. A key point in solving rational equations is to ensure that no term is divided by zero, which often requires checking any proposed solutions.

Once simplified to a simpler algebraic equation (often polynomial), we can proceed to solve it using standard algebra techniques.

Quadratic Formula

Once the rational equation is cleared of denominators, it often leads to a polynomial equation. In our solution, it reduced to a quadratic equation: \(r^2 - 15r - 250 = 0\). Quadratic equations are of the form \(ax^2 + bx + c = 0\).

The quadratic formula provides a method to find the roots of any quadratic equation and is given by: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our specific equation:

• \(a = 1\)
• \(b = -15\)
• \(c = -250\)

Substituting these values into the quadratic formula allows us to solve for \(r\). This yields two potential solutions because of the \(\pm\) in the formula. It means we calculate once for the addition, and once for the subtraction of the square root term.

The quadratic formula is reliable for solving quadratics, especially when factorization appears complicated or impossible using simple methods.

Checking Solutions in Algebra

Checking solutions is crucial, especially in rational equations. The main reason is to ensure the solution doesn't make any original denominator equal to zero, which would be undefined.

To check the solutions:

• Substitute each solution back into the original equation.
• Evaluate whether both sides of the original equation balance.
• Confirm that none of the denominators are zero with the given solution.

In our problem, once solutions for \(r\) are found, both should be substituted back into the initial equation \(\frac{5}{3} + \frac{250}{9r} = \frac{r}{9}\) to verify their validity.

Only the solutions that do not make any denominator zero are valid. This step is ultimately about confirming the integrity of your solutions and ensuring they make sense in the original context of the problem.