Problem 25 - Algebra and Trigonometry

Question

\(9- 46\) The given equation is either linear or equivalent to a linear equation. Solve the equation. $$ \frac{2}{t+6}=\frac{3}{t-1} $$

Step-by-Step Solution

Answer

The solution is \( t = -20 \).

Step 1: Identify the equation type

The given equation is of the form \( \frac{2}{t+6} = \frac{3}{t-1} \). This is a rational equation, but we can solve it as a linear equation by eliminating the fractions.

Step 2: Cross-multiply to eliminate fractions

Cross-multiply to get rid of the fractions: \[ 2(t - 1) = 3(t + 6) \]

Step 3: Expand both sides of the equation

Distribute the constants across each term within the parentheses: \[ 2t - 2 = 3t + 18 \]

Step 4: Rearrange to isolate terms with t

Move all terms involving \(t\) to one side of the equation: Subtract \(2t\) from both sides, \[ -2 = t + 18 \]

Step 5: Isolate the variable t

Subtract 18 from both sides to solve for \(t\), \[ -2 - 18 = t \] Thus, \[ t = -20 \]

Step 6: Verify the solution

Substitute \(t = -20\) back into the original equation to check: Left side: \(\frac{2}{-20+6} = \frac{2}{-14} = -\frac{1}{7}\) Right side: \(\frac{3}{-20-1} = \frac{3}{-21} = -\frac{1}{7}\) Both sides are equal, hence verified.

Key Concepts

Rational Equations

Rational equations are mathematical expressions in which two rational expressions, also known as fractions, are set equal to each other. Each rational expression usually consists of a numerator and a denominator that are polynomials. Solving rational equations often involves finding the variable value that makes the expression true. It's crucial to understand that solutions must not make any denominators equal to zero, as division by zero is undefined.
In the exercise provided, we worked with the rational equation \( \frac{2}{t+6} = \frac{3}{t-1} \). Rational equations require attention to the denominators. Before proceeding with solving them, we make a note that neither \(t+6\) nor \(t-1\) should be zero, determining the values of \(t\) that would make them zero and excluding those from our solution.
In this case, \(t = -6\) and \(t = 1\) are not valid solutions because they would make the denominators zero.

Cross-Multiplication

Cross-multiplication is a valuable technique for solving equations involving fractions. When presented with two fractions set equal to each other, such as \( \frac{a}{b} = \frac{c}{d} \), cross-multiplication allows us to eliminate the fractions by multiplying across the equals sign. This means multiplying the numerator of one fraction by the denominator of the other and vice versa.
The provided exercise crossed-multiplies to eliminate fractions:

• The equation \( \frac{2}{t+6} = \frac{3}{t-1} \) becomes \( 2(t-1) = 3(t+6) \) after cross-multiplication.

This step helps in transforming the equation into a more manageable form without the complexity of fractions, paving the way for straightforward algebraic manipulation.

Algebraic Manipulation

Algebraic manipulation involves a series of operations or transformations applied to an equation to simplify or solve it. It often includes expanding expressions, combining like terms, or arranging terms in a favorable order.
Once cross-multiplication eliminates the fractions, the next step is algebraic manipulation of the equation's components. From the previous step, we transform \( 2(t-1) = 3(t+6) \) into:

• Distribute 2 across \((t-1)\) to get \( 2t - 2 \).
• Distribute 3 across \((t+6)\) to get \( 3t + 18 \).

The equation becomes \( 2t - 2 = 3t + 18 \). These operations are crucial to balance and simplify both sides for further variable isolation.

Solving for a Variable

Solving for a variable means rearranging an equation to isolate the variable on one side. It involves a series of steps to move terms from one side of the equation to the other, usually following the order of operations in reverse.
In the final steps of the given exercise, we solved for the variable \(t\) in the transformed equation \( 2t - 2 = 3t + 18 \). Here's how you proceed:

• Rearrange to group terms with \(t\) on one side: Subtract \(2t\) from both sides to get \(-2 = t + 18\).
• Next, subtract 18 from both sides to isolate \(t\), resulting in \(-2 - 18 = t\), which simplifies to \(t = -20\).

After isolating \(t\), verify the solution by plugging it back into the original equation to ensure both sides equal, confirming \( t = -20 \) satisfies the equation without causing division by zero.