Problem 35 - Algebra 1

Question

Solve the proportion. Check for extraneous solutions. $$\frac{x-3}{x}=\frac{x}{x+6}$$

Step-by-Step Solution

Answer

The solution of the given proportion is \(x = 6\).

Step 1: Cross Multiplication

Cross-multiply to eliminate the fractions. Resulting equation is: \((x-3)*(x+6) = x*x\).

Step 2: Simplification

Expand and simplify the equation: \(x^2 - 3x + 6x - 18 = x^2\). Combine like terms to get: \(x^2 + 3x - 18 = x^2\). Subtract \(x^2\) from both sides of the equation, we have: \(3x - 18 = 0\).

Step 3: Solve for x

Add 18 to both sides of the equation and divide by 3: \(x = 6\). It's the probable solution.

Step 4: Check for Extraneous Solution

Substitute x = 6 into the original equation to check for extraneous solutions. Substitution results in \(3/6 = 6/12\), which simplifies to \(1/2 = 1/2\). Since it is a valid equality, x = 6 is the required solution and not an extraneous solution.

Key Concepts

Cross Multiplication in Proportions

When we are faced with an equation that is a proportion involving fractions, **cross multiplication** is a handy technique to simplify the solution process. In essence, cross multiplication helps in getting rid of the fractions so we can deal with a simple equation.

Here's how it works:

• For an equation of the form \(\frac{a}{b} = \frac{c}{d}\), cross multiplication means multiplying \(a\) by \(d\) and \(b\) by \(c\). This gives us the equation: \(ad = bc\).
• Applied to our original equation \(\frac{x-3}{x} = \frac{x}{x+6}\), cross multiplication leads us to: \((x-3) \cdot (x+6) = x \cdot x\).

After this step, we move on to simplifying the resulting equation. Cross multiplication is particularly crucial because it transforms a complicated proportion into a simpler polynomial equation by eliminating the denominators.

Understanding Extraneous Solutions

In mathematics, when we solve equations, especially those involving fractions or square roots, we sometimes encounter **extraneous solutions**. These are solutions that emerge in the process of solving but are not valid when substituted back into the original equation.

The importance of identifying extraneous solutions lies in ensuring the correctness of your final answer. Let's look at how to check for them:

• Once you solve for \(x\), substitute it back into the original equations.
• If the values simplify correctly (for example \( \frac{3}{6} = \frac{6}{12} \)), your solution is valid.
• If not, then the solution is extraneous.

In our exercise, substituting \(x = 6\) gives us \(\frac{3}{6} = \frac{6}{12}\), which simplifies to \(\frac{1}{2} = \frac{1}{2}\). This confirms that \(x = 6\) is not an extraneous solution, affirming it as the valid answer.

Simplifying Equations for Solution

Simplifying equations is a key step in solving mathematical problems, converting complex forms into simpler ones that are easier to solve.

Here’s how we simplify:

• Expand any expressions, such as using the distributive property where necessary. In our case, \((x-3)(x+6) = x^2 - 3x + 6x - 18\).
• Combine like terms, making the equation simpler: \(x^2 - 3x + 6x - 18 = x^2\) simplifies further to \(x^2 + 3x - 18 = x^2\).
• Isolate the unknown variable by performing operations such as adding or subtracting terms from both sides and dividing, when needed. We subtract \(x^2\) from both sides, leading to \(3x - 18 = 0\).

Finally, solve for the variable using operations to isolate it. For example, add 18 to both sides and divide by 3, resulting in \(x = 6\). Simplification transforms the equation into a form that directly gives us the solution.