Problem 27 - Algebra 2

Question

Solve each equation or inequality. Check your solutions. $$ \frac{4}{z-2}-\frac{z+6}{z+1}=1 $$

Step-by-Step Solution

Answer

The solutions are \( z = \frac{1 \pm \sqrt{145}}{4} \), and both are valid.

Step 1: Find a Common Denominator

The given expression is \( \frac{4}{z-2} - \frac{z+6}{z+1} = 1 \). The least common denominator of these fractions is \((z-2)(z+1)\). Multiply each term by \((z-2)(z+1)\) to eliminate the fractions, resulting in:\( 4(z+1) - (z+6)(z-2) = (z-2)(z+1) \).

Step 2: Expand and Simplify

Expand each term in the equation:\( 4(z+1) = 4z + 4 \)\( (z+6)(z-2) = z^2 - 2z + 6z - 12 = z^2 + 4z - 12 \)\( (z-2)(z+1) = z^2 - 2 + z - 2 = z^2 - z - 2 \)The equation becomes:\( 4z + 4 - (z^2 + 4z - 12) = z^2 - z - 2 \).

Step 3: Combine Like Terms

Combine like terms on the left side:\( 4z + 4 - z^2 - 4z + 12 = z^2 - z - 2 \)Simplifying gives:\( -z^2 + 16 = z^2 - z - 2 \).

Step 4: Move All Terms to One Side

Move all terms to the left side of the equation:\( -z^2 + 16 - z^2 + z + 2 = 0 \)This simplifies to:\( -2z^2 + z + 18 = 0 \).

Step 5: Simplify Further

As we simplify further:Multiplying the whole equation by -1 (to make the leading term positive) gives us:\( 2z^2 - z - 18 = 0 \). Now, solve this quadratic equation using the quadratic formula.

Step 6: Use the Quadratic Formula

Use the quadratic formula, \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2, b = -1, c = -18 \).Calculate the discriminant: \( b^2 - 4ac = (-1)^2 - 4(2)(-18) = 1 + 144 = 145 \).The solutions are:\( z = \frac{-(-1) \pm \sqrt{145}}{2 \times 2} \).This simplifies to:\( z = \frac{1 \pm \sqrt{145}}{4} \).

Step 7: Check the Solutions

For both solutions, check that they do not make any denominator zero in the original equation. Since neither solution equals 2 or -1, substitute both values back into the original equation to verify they satisfy it.

Key Concepts

Common Denominator

When solving equations and inequalities with fractions, finding a common denominator is a crucial first step. This allows us to combine and manipulate the fractions easily. In the exercise, we began with the expression \( \frac{4}{z-2} - \frac{z+6}{z+1} = 1 \). Here, each fraction has a different denominator, \(z-2\) and \(z+1\). To eliminate the fractions, we need to find the least common denominator (LCD), which is \((z-2)(z+1)\). This means multiplying every term of the equation by the LCD. By doing this, we clear the fractions and simplify our task to a straightforward polynomial equation. Finding a common denominator helps in:

• Eliminating fractions from the equation
• Simplifying the expression for easier manipulation
• Ensuring the equation maintains its balance when cleared of fractions

This step sets the stage for easier manipulation of the equation as seen in the subsequent steps.

Quadratic Formula

Once we have a quadratic equation, like the one derived in this problem \(2z^2 - z - 18 = 0\), we can use the quadratic formula to find the solutions. The quadratic formula is a powerful tool for solving any quadratic equation and is given by:\[z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \(a\), \(b\), and \(c\) are the coefficients from the equation \(ax^2 + bx + c = 0\). In our case, \(a = 2\), \(b = -1\), and \(c = -18\). The first thing we do is calculate the discriminant, \(b^2 - 4ac\), which tells us the nature of the roots:

• If it is positive, there are two distinct real solutions.
• If it is zero, there is exactly one real solution (a repeated root).
• If it is negative, the solutions are complex numbers (not real).

For this exercise, the discriminant is calculated as \(145\), which is positive, thus indicating two distinct real solutions. Applying the quadratic formula, we find:\[z = \frac{1 \pm \sqrt{145}}{4}\] This tells us the values of \(z\) that satisfy the quadratic equation derived from the original problem.

Combine Like Terms

Combining like terms is a fundamental aspect of simplifying equations. In the step-by-step solution, after multiplying by the common denominator, we have an expanded equation that looks complex. For example, we had: \[-2z^2 + z + 18 = 0\]When combining like terms, we group together similar variables and constants to make the equation simpler and more manageable. This typically involves:

• Identifying all terms with the same variable and degree, for example, putting all \(z^2\) terms together.
• Adding or subtracting the coefficients of these terms.
• Simplifying constant terms likewise by adding or subtracting them.

In this exercise, combining terms led to moving all expressions to the left side and simplifying them, which eventually gave us the clean quadratic form necessary for using the quadratic formula. Simplification through combining terms not only makes the equation easier to solve but also lessens the chance of errors while advancing to more complex steps, like applying the quadratic formula.

Check Solutions

Verifying the solutions is a must-do step when solving equations and inequalities. After finding solutions using methods like the quadratic formula, we must ensure these solutions are valid by checking them. For this particular exercise, we found two potential solutions for \(z\):\[\frac{1 \pm \sqrt{145}}{4}\]The importance of checking lies in ensuring that these solutions don't create undefined situations, such as zero denominators, in the original equation.

• Substitute each solution back into the original equation.
• Make sure that no division by zero occurs.
• Verify that the left-hand side equals the right-hand side of the equation when the solution is plugged back in.

If any of the solutions do not satisfy the original equation, they should be discarded as extraneous solutions. In this exercise, after checking the solutions, we confirmed none of them made the originally provided denominators zero, ensuring the solutions were valid and complete.