Problem 74 - Intermediate Algebra

Question

Solve equation. If a solution is extraneous, so indicate. \(\frac{z+2}{z+8}=\frac{z-3}{z-2}\)

Step-by-Step Solution

Answer

The solution is \(z = 4\), and it is not extraneous.

Step 1: Cross-Multiply to Eliminate Fractions

Start by cross-multiplying the terms in the equation: \[ (z + 2)(z - 2) = (z - 3)(z + 8) \]This step is necessary to get rid of the fractions and simplify the equation.

Step 2: Expand Both Sides of the Equation

Expand both sides of the equation using the distributive property:\[ (z + 2)(z - 2) = z^2 - 2z + 2z - 4 = z^2 - 4 \]\[ (z - 3)(z + 8) = z^2 + 8z - 3z - 24 = z^2 + 5z - 24 \]

Step 3: Set the Equation to Zero

Now set the two expressions equal to each other and subtract one side from the other to set the equation to zero:\[ z^2 - 4 = z^2 + 5z - 24 \]Subtract \(z^2\) and \(-4\) from both sides:\[ 0 = 5z - 20 \]

Step 4: Solve for z

Isolate \(z\) by adding or subtracting terms:\[ 5z = 20 \]Now divide each side by 5:\[ z = 4 \]

Step 5: Check for Extraneous Solutions

Check if the solution makes any of the denominators in the original equation zero. Substitute \(z = 4\) back into the denominators:- \(z + 8 = 12\)- \(z - 2 = 2\)Since none of these denominators are zero, \(z = 4\) is not an extraneous solution and is valid.

Key Concepts

Cross-Multiplication

Cross-multiplication is a crucial technique used to solve equations that contain fractions, especially those with variables in the denominator. Imagine you have two fractions set equal to each other, like in the original exercise: \( \frac{z+2}{z+8} = \frac{z-3}{z-2} \). The goal here is to eliminate the fractions and make the equation easier to handle. We achieve this by cross-multiplying. The term "cross-multiply" refers to multiplying the numerator of one fraction by the denominator of the other fraction and doing the same in reverse.For this exercise:

• Multiply \((z + 2)\) (the numerator on the left side) by \((z - 2)\) (the denominator on the right side).
• Multiply \((z - 3)\) (the numerator on the right side) by \((z + 8)\) (the denominator on the left side).

This gives us: \[ (z + 2)(z - 2) = (z - 3)(z + 8) \].Cross-multiplication simplifies an equation with several terms into one that is usually easier to solve. It transforms a proportion or fraction equation into a polynomial equation, allowing us to use other algebraic techniques to solve for the variable.

Distributive Property

The distributive property is a fundamental algebraic concept used to expand expressions. It's often phrased as "distributing" the multiplication over addition or subtraction. Mathematically, the distributive property states that for any numbers \(a, b,\) and \(c\), we have:\[ a(b + c) = ab + ac \].This property is essential when we multiply binomials, as seen in the original solution.In step 2 of the solution, the problem required us to expand both sides of the equation using the distributive property:

• For \((z + 2)(z - 2)\), distribute each term:
  • \(z \cdot z = z^2\)
  • \(z \cdot -2 = -2z\)
  • \(2 \cdot z = 2z\)
  • \(2 \cdot -2 = -4\)
  Combine these to get \(z^2 - 4\), as the middle terms cancel out.
• For \((z - 3)(z + 8)\), apply similar steps:
  • \(z \cdot z = z^2\)
  • \(z \cdot 8 = 8z\)
  • \(-3 \cdot z = -3z\)
  • \(-3 \cdot 8 = -24\)
  Combine these to form \(z^2 + 5z - 24\).

The distributive property is invaluable for simplifying equations before solving them!

Extraneous Solution

An extraneous solution is a solution that emerges from the algebraic manipulations of an equation but doesn't actually satisfy the original equation. Such solutions often arise when we perform operations that are not reversible, like squaring both sides of an equation or cross-multiplying fractions.In our problem, after solving the equation, we obtained the solution \(z = 4\). However, it's vital to verify that this solution does not cause any denominators in the original equation to be zero, because that would make the original equation undefined.To check if \(z = 4\) is extraneous, substitute it back into the denominators of the original equation:

• For \(z + 8\), substitute: \(4 + 8 = 12\), which is not zero.
• For \(z - 2\), substitute: \(4 - 2 = 2\), which is not zero.

Since neither denominator becomes zero, \(z = 4\) is a valid solution and not extraneous. Always remember to check for extraneous solutions by substituting back into the original equation to ensure validity.