Problem 14 - Algebra 1: Concepts and Skills

Question

Solve the equation. Check your solution in the original equation. $$ \frac{4}{5} t=0 $$

Step-by-Step Solution

Answer

The solution to the equation is \( r = 50 \).

Step 1: Distribute

First, distribute terms in the equation to get rid of parentheses: \( 8 * 4 - 8 * r + r = -6 * 3 - 6*r \). This gives: \( 32 - 8r + r = -18 - 6r \).

Step 2: Combine Like Terms

Next, combine the like terms on both sides of the equation. This results in \( 32 - 7r = -18 - 6r \).

Step 3: Isolate the Variable r

Starting by adding \( 7r \) to both sides of the equation, we have \( 32 = -18 + r \). Then by adding 18 to both sides, the variable r is isolated on the right side, so the equation becomes \( r = 50 \).

Step 4: Check the Solution

Substitute \( r = 50 \) into the original equation to check the solution. It is important to note that both sides of the equation should have the same value if the solution is correct. Thus, the left side becomes \( 8(4-50) + 50 = -392 \), and the right side becomes \( -6(3+50) = -392\). As both sides are equal, the solution is correct.

Key Concepts

Distribution in Algebra

In algebra, distribution refers to the process of distributing a number outside of a parenthesis to each term inside the parenthesis. This is also known as the distributive property. It's an essential tool when you want to simplify expressions and solve equations.
For the equation given in the exercise:

• We used distribution to handle the terms inside the parentheses: \(8(4-r)+r=-6(3+r)\)
• The distributive property tells us to multiply each term inside the parentheses by the number outside. This leads to: \(8 \times 4 - 8 \times r + r = -6 \times 3 - 6 \times r\).
• Simplifying further gives us: \(32 - 8r + r = -18 - 6r\).

By clearing out the parentheses, we moved one step closer to isolating the variable. Always remember that every term inside the parenthesis gets multiplied by the term outside for accurate distribution.

Like Terms

Like terms are terms in an algebraic expression that have the same variables raised to the same powers. These terms can be combined, similar to how you would add numbers.

• In our equation, after distributing, we had \(32 - 8r + r = -18 - 6r\).
• The like terms here are the terms with the variable \(r\): \(-8r\) and \(+r\) on the left,
  and \(-6r\) on the right.
• Combining the like terms leads to: \(32 - 7r = -18 - 6r\).

Combining like terms simplifies equations and helps move towards isolating the variable. As you practice, identifying and combining these terms will become second nature.

Variable Isolation

Variable isolation involves rearranging an equation so that the variable you're solving for is on one side of the equation, and everything else is on the other. This is a critical step to find the value of the variable.

• We started with: \(32 - 7r = -18 - 6r\)
• The goal was to isolate \(r\). First, add \(7r\) to both sides, resulting in: \(32 = -18 + r\).
• Next, add 18 to both sides to fully isolate \(r\), giving us: \(r = 50\).

Through gradual rearrangement, you can uncover the solution step by step, even in more complex equations. This systematic method helps maintain accuracy and clarity in solving algebraic equations.

Solution Checking

After solving an equation, checking your solution is a crucial step that confirms the answer is correct. By substituting the calculated value into the original equation, we verify if both sides of the equation are equal.

• For our equation, substitute \(r = 50\) back: \(8(4-50) + 50 = -392\) and \(-6(3+50) = -392\).
• Both calculations lead to \(-392\), confirming that \(r = 50\) is indeed correct.

Verification prevents mistakes and ensures your solution is valid. Whether you're working through homework or real-world problems, regularly check your answers to build confidence in your problem-solving abilities.