Problem 18 - Algebra 1: Concepts and Skills

Question

Solve the equation. Check your solution in the original equation. $$ \frac{1}{5}(25-5 k)=21-3(k-4) $$

Step-by-Step Solution

Answer

The solution to the equation is \( k = 33 / 4 \)

Step 1: Distribute coefficients to the bracketed terms

Begin by applying the distributive property to multiply the fractions by the bracketed terms. This gives: \( \frac{1}{5}*25 - \frac{1}{5}*5k = 21 - 3*k +3*4 \)

Step 2: Simplify both sides of the equation

Next simplify the equation to: \( 5 - k = 21 - 3k + 12 \)

Step 3: Bring like terms together

Rearrange the equation to gather like terms on one side of the equation: \( 5 + 3k - 21 = -k +12 \)

Step 4: Simplify the equation

Now simplify the equation to: \( 4k = 33 \)

Step 5: Solve for 'k'

To solve for 'k', divide both sides of the equation by 4. The solution is: \( k = 33 / 4 \)

Step 6: Check your solution in the original equation

Substitute \( k = 33 / 4 \) back into the original equation to verify the solution. If both sides of the equation are equal, then your solution is correct.

Key Concepts

Distributive Property

Understanding the distributive property is crucial when dealing with algebraic expressions that involve parentheses. Essentially, it allows us to multiply a single term by each term inside a set of parentheses. For example, if you see an expression like \( a(b+c) \), it's the distributive property that enables you to rewrite this as \( ab + ac \).

Let's apply this to our exercise: We're starting with \(\frac{1}{5}(25-5k)=21-3(k-4)\). By distributing the \(\frac{1}{5}\) across \(25\) and \( -5k\), and the \( -3\) across \(k\) and \( -4\), we're breaking the equation down into smaller parts, which are easier to handle. This step is the foundation that allows us to move forward in solving the equation.

Simplifying Equations

Once the distributive property is applied, the next step is simplifying the equation. This involves combining like terms and performing basic arithmetic operations to get the equation to a more manageable state. It's like cleaning up a messy room so you can clearly see what you have to work with.

In our exercise, after distribution, we simplify \( 5 - k = 21 - 3k + 12 \). This means we add or subtract all the constant terms (the numbers without variables) and combine the variable terms (the 'k' terms). Simplifying makes equations less cluttered and brings us one step closer to isolating the variable, which is our ultimate goal in solving for 'k'. Remember to perform the same operations on both sides of the equation to maintain balance.

Variable Isolation

The final aim in solving a linear equation is to isolate the variable—getting 'k' by itself on one side of the equation. This is achieved by moving terms that don't contain the variable to the opposite side. Think of it like isolating a character in a play to spotlight their solo—here, 'k' is our star, and we want it to have the entire stage.

In step 3, we started this process by rearranging the equation to \( 3k + 5 - 21 = -k + 12 \). We then continued to simplify to \( 4k = 33 \), which provided a clear path to hone in on 'k'. By dividing both sides by 4, we isolate 'k' and find that \( k = \frac{33}{4} \). The path to variable isolation involves smart rearrangement and consistent simplification. These strategies helmed us solve for 'k' effectively in our example.