Problem 103 - Intermediate Algebra

Question

Solve each equation. $$ \frac{t-2}{5}+5 t=\frac{7}{5}-\frac{t-2}{2} $$

Step-by-Step Solution

Answer

The solution is \( t = \frac{28}{57} \).

Step 1: Align the Equation

Start by writing the given equation:\[ \frac{t-2}{5} + 5t = \frac{7}{5} - \frac{t-2}{2} \]

Step 2: Eliminate Fractions

To make it easier to solve the equation, eliminate the fractions by finding the least common multiple (LCM) of the denominators, which is 10.Multiply every term by 10 to remove the fractions:\[ 10 \left( \frac{t-2}{5} \right) + 10(5t) = 10 \left( \frac{7}{5} \right) - 10 \left( \frac{t-2}{2} \right) \]

Step 3: Simplify Each Term

Solve the multiplication of each part:\[ 2(t-2) + 50t = 2(7) - 5(t-2) \]Simplify each term:\[ 2t - 4 + 50t = 14 - 5t + 10 \]

Step 4: Combine Like Terms

Combine like terms on both sides of the equation:\[ (2t + 50t) - 4 = (14 + 10) - 5t \]This becomes:\[ 52t - 4 = 24 - 5t \]

Step 5: Isolate the Variable

Move all terms involving \( t \) to one side and constants to the other:\[ 52t + 5t = 24 + 4 \]This simplifies to:\[ 57t = 28 \]

Step 6: Solve for t

Divide both sides by 57 to solve for \( t \):\[ t = \frac{28}{57} \]Simplifying \( \frac{28}{57} \), we find it cannot be reduced further, so:\[ t = \frac{28}{57} \]

Key Concepts

Fraction Elimination

Fractions in equations can make solving them tricky. Elimination is a helpful first step to simplify things. Think of fractions as little barriers you might want to get rid of right away. To eliminate fractions, first identify the denominators in the equation. For our equation, we have denominators of 5 and 2.

Here's the trick: find a common denominator which allows you to clear all fractions at once! An easy way to do this is by using the least common multiple (LCM). In this case, the LCM of 5 and 2 is 10. By multiplying every term in the equation by 10, we effectively eliminate all fractions.

• Consider the term \( \frac{t-2}{5} \). Multiplying it by 10 gives \( 10 \times \frac{t-2}{5} = 2(t-2) \).
• Do this for each term involving a fraction, making sure to multiply each term accurately.

This turns the equation into something much simpler and easier to manage!

Least Common Multiple

The least common multiple (LCM) is a handy concept when dealing with numbers like in fractions. The LCM of two numbers is the smallest number that is a multiple of both. Finding the LCM lets us remove fractions from the equation quickly.

For example, consider the fraction denominators 5 and 2. What number fits inside both of these? It's 10! That's the smallest number they both divide into without a remainder, hence the LCM.

• 5 fits into 10 because 5 x 2 = 10.
• 2 fits into 10 because 2 x 5 = 10.

We use this number to scale up elements in the equation, getting rid of fractions entirely. It keeps things clean and paves the way for smooth sailing through the rest of the solution.

Combining Like Terms

Combining like terms is essential to solving linear equations efficiently. Think of like terms as ingredients of the same kind that you can mix. They have the same variable parts, which makes them stackable.

When you simplify an equation, it's important to look at all the "like" parts. In our example, after removing fractions, you’re left with something like \( 2t - 4 + 50t \). Here, the like terms are \( 2t \) and \( 50t \).

• Add the coefficients: \( 2 + 50 = 52 \).
• Your combined term is \( 52t \).

Combining like terms brings clarity and simplicity to equations, making subsequent steps clearer and more straightforward.

Isolating the Variable

Isolating the variable is one of the final and most critical steps in solving an equation. Here, your goal is to get the variable, in this case, \( t \), all by itself on one side of the equation.

After combining like terms, you might have something like \( 52t - 4 = 24 - 5t \). To isolate \( t \), follow these steps:

• Move all terms involving \( t \) to one side: Add \( 5t \) to both sides, getting \( 52t + 5t = 24 + 4 \).
• Combine these to get \( 57t = 28 \).
• Finally, divide both sides by 57 to solve for \( t \): \( t = \frac{28}{57} \).

Isolating the variable lets you see exactly what \( t \) is equated to, giving you the solution in its simplest possible form.