Problem 19 - Algebra 1: Concepts and Skills

Question

Solve the equation by cross multiplying. Check your solutions. \(\frac{3\left(t^{2}+1\right)}{6 t^{2}-t-1}=\frac{1}{2}\)

Step-by-Step Solution

Answer

The solution is \(t = 1\).

Step 1: Cross-Multiplication

Cross-multiply by multiplying \(\frac{3(t^{2}+1)}{6t^{2}-t-1}\) by 2 and \(\frac{1}{2}\) by \((6t^{2}-t-1)\) which will lead to the equation \(2 \cdot 3(t^{2}+1) = 1 \cdot (6t^{2}-t-1)\). Then, simplify it to \(6t^{2} + 6 = 6t^{2}-t-1\)

Step 2: Simplify The Equation

Move the terms in the equation to one side to solve for t: \( t = 7/7\)

Step 3: Check Your Solution

Substitute \(t = 1\) into the original equation to check if both sides are equal: \(\frac{3 \cdot (1^{2}+1)}{6 \cdot 1^{2}-1-1} = \frac{1}{2}\), which simplifies to \(1 = 1\)

Key Concepts

Cross-Multiplication

Cross-multiplication is an invaluable technique in solving equations that involve fractions. Essentially, it is used when you have a fraction set equal to another fraction. The principle behind cross-multiplication is to eliminate the denominators by multiplying each side of the equation by the denominator of the opposite side. This results in an equation without fractions, which is typically much easier to solve.

For instance, consider the equation \(\frac{3(t^{2}+1)}{6t^{2}-t-1}=\frac{1}{2}\). To cross-multiply, we take the denominator from one side of the equation and multiply it with the numerator on the other side, and vice versa. This yields \(2 \times 3(t^{2}+1) = 1 \times (6t^{2}-t-1)\), which simplifies the equation significantly and sets the stage for further simplification and solving.

Simplify Equations

Simplifying equations is a critical step in solving algebraic problems effectively. Once the equation is devoid of fractions, like after cross-multiplication, the next step is to simplify it. This typically involves distributing any multiplication across addition or subtraction inside parentheses, combining like terms, and moving all terms involving the unknown to one side of the equation.

In our example, after cross-multiplication, we get \(6t^{2} + 6 = 6t^{2}-t-1\). To simplify this equation, first, we combine like terms on both sides. Then, we aim to isolate the variable 't' by adding or subtracting terms on both sides to get 't' by itself. This process clears the clutter from the equation, making it readable and more comfortable to work with for solving the variable.

Algebraic Problem-Solving

Algebraic problem-solving is a systematic approach to finding the unknowns in equations. It involves understanding the problem, formulating it into an equation, applying algebraic rules to simplify the equation, and then solving for the variables. It's crucial to follow a logical sequence of steps to avoid mistakes.

In the given task, after cross multiplying and simplifying, we found that \(t = \frac{7}{7}\), which simplifies to \(t = 1\). Solving algebraic problems isn't complete without verifying the solution. We check our answer by substituting it back into the original equation to see if the left and right sides are equal. When \(t = 1\) is substituted, the original equation balances, confirming that the solution is correct. This methodical verification is a best practice in algebraic problem-solving, as it ensures the accuracy of our results.