Problem 31 - Algebra and Trigonometry

Question

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 to the equation is \(t = -20\).

Step 1: Clear the Fractions

To eliminate the fractions, multiply both sides of the equation by the product of the denominators, \((t+6)(t-1)\). This gives:\[(t+6)(t-1) \cdot \frac{2}{t+6} = (t+6)(t-1) \cdot \frac{3}{t-1}\]Simplifying both sides, we get:\[2(t-1) = 3(t+6)\]

Step 2: Expand Both Sides

Distribute the numbers outside the parentheses:\[2 \cdot t - 2 \cdot 1 = 3 \cdot t + 3 \cdot 6\]Simplifying both sides we have:\[2t - 2 = 3t + 18\]

Step 3: Rearrange to Isolate Variable

Subtract \(2t\) from both sides to move terms involving \(t\) to one side:\[2t - 2 - 2t = 3t + 18 - 2t\]This simplifies to:\[-2 = t + 18\]

Step 4: Solve for the Variable

Subtract 18 from both sides to isolate \(t\):\[-2 - 18 = t\]Thus, we find:\[t = -20\]

Step 5: Verify the Solution

Substitute \(t = -20\) back into the original equation to check:\[\frac{2}{-20 + 6} = \frac{3}{-20 - 1}\]This simplifies to:\[\frac{2}{-14} = \frac{3}{-21}\]Both sides reduce to \(-\frac{1}{7}\), confirming \(t = -20\) satisfies the equation.

Key Concepts

Clearing Fractions

When solving equations that include fractions, the goal is often to eliminate the fractions to simplify the equation. This process is known as "clearing fractions." To do this, you multiply each term in the equation by a common denominator that is the product of each fraction's denominator. In the original problem, this means multiplying both sides by

• (t+6)(t-1)

This step cancels out the denominators, leaving behind an equation that is easier to work with. For instance, after multiplying, we transition from:

• \(\frac{2}{t+6} = \frac{3}{t-1}\)

to:

• \[2(t-1) = 3(t+6)\]

This new equation is free of fractions, preparing the ground for the next steps. Clearing fractions streamlines problems and makes calculations more straightforward, reducing the chance of errors in subsequent steps.

Distributing Terms

Once the fractions are cleared, the next focus is on "distributing." This means applying distribution to multiply out any terms next to parentheses. Each term outside the parenthesis must be multiplied by every term inside the parenthesis. Using the equation:

• \[2(t-1) = 3(t+6)\]

Distribution assists in expanding the equation. You perform multiplication as follows:

• 2 times \(t\)
• 2 times -1
• 3 times \(t\)
• 3 times 6

This builds the expression:

• \[2t - 2 = 3t + 18\]

Distributing terms opens up the equation, facilitating the rearranging and simplifying processes that are to come. Ensure each term is correctly expanded to avoid mistakes as you move toward solving the equation.

Isolating Variables

The ultimate goal in solving linear equations is to determine the value of the variable. This requires "isolating" the variable on one side of the equation. Starting from:

• \[2t - 2 = 3t + 18\]

The first step to isolate the variable is to consolidate all terms involving the variable to one side of the equation:

• Subtract \(2t\) from both sides to remove \(t\) terms from one side

This gives:

• \[-2 = t + 18\]

Next, eliminate constants from the side with the variable by subtracting 18 from both sides:

• \[-2 - 18 = t\]

Thus, you'll find:

• \[t = -20\]

Isolating the variable often requires rearranging and simplifying the equation step by step until the variable is alone. Always perform these steps with care to ensure accurate solutions. The final check is substituting back to verify your solution's correctness.