Problem 56 - Elementary Algebra

Question

Solve the equation \(6[2(x-4)+1]=3[2(x-7)]\).

Step-by-Step Solution

Answer

Answer: The value of x in the given equation is 0.

Step 1: Apply the distributive property

Start by applying the distributive property to both expressions inside the brackets. \(6[2(x-4)+1] = 6(2x - 8 + 1)\) and \(3[2(x-7)] = 3(2x - 14)\). Notice that we multiply the number on the outside of the bracket to the terms inside the bracket for both expressions.

Step 2: Simplify expressions

Next, simplify the expressions in the equation: \(6(2x - 7) = 3(2x - 14)\). Now, perform the multiplication: \(12x - 42 = 6x - 42\).

Step 3: Subtract '6x' from both sides

To isolate the variable x, subtract '6x' from both sides of the equation: \((12x - 6x) - 42 = 6x - 6x - 42\). This simplifies to \(6x - 42 = -42\).

Step 4: Add '42' to both sides

To further isolate the variable x, add '42' to both sides of the equation: \(6x - 42 + 42 = -42 + 42\). This simplifies to \(6x = 0\).

Step 5: Divide by '6'

Now, to find the value of x, divide both sides of the equation by '6': \(\frac{6x}{6} = \frac{0}{6}\). This gives us the final answer: \(x = 0\).

Key Concepts

Distributive Property

When we come across problems that involve parentheses, the distributive property is the tool we need to simplify them. This property states that multiplying a sum by a number is the same as multiplying each addend by the number and then adding the products. In essence, it involves spreading out the multiplication across the terms within the parentheses. For example, if we have \(a(b + c)\), applying the distributive property would give us \(ab + ac\).

Let's apply this to our exercise. We have \(6[2(x-4)+1]\) and \(3[2(x-7)]\). Using the distributive property, we multiply \(6\) by each term inside the first bracket and \(3\) by each term inside the second bracket. This step is crucial because it transforms the equation into a more manageable form without parentheses, making it easier to solve.

Isolate Variable

To solve for a variable means to isolate that variable on one side of the equation. It's all about performing operations that will get the variable by itself, thus 'isolating' it. The goal here is to have the variable (let's call it \(x\)) on one side and the numbers on the other, so we can clearly see what \(x\) equals.

In our exercise, we wanted to isolate \(x\) in the equation \(12x - 42 = 6x - 42\). To do this, we subtract \(6x\) from both sides to get rid of the \(x\)-term on the right side. Now \(x\) appears only on the left side, and we're closer to finding its value. Remember: whatever you do to one side of the equation, you must do to the other to maintain balance.

Simplify Expressions

Simplifying expressions is a bit like tidying up a messy room; we want to make the equation as neat and as straightforward as possible. This process can involve combining like terms (terms that have the same variable raised to the same power), canceling out identical terms on both sides, and performing basic arithmetic operations.

In our equation, after applying the distributive property, we simplified the expression \(6(2x-7)\) to \(12x-42\). During the steps to isolate the variable, we also simplified by canceling terms on both sides of the equation. Simplifying an expression makes it easier to see how close the variable is to being isolated and ultimately solved.

Equation Solving Steps

Solving linear equations is a methodical process. Here are the general steps to follow:

• Start with simplifying both sides of the equation as much as possible.
• Use the distributive property to eliminate parentheses.
• Combine like terms on each side.
• Get all variable terms on one side and numbers on the other by adding or subtracting.
• Multiply or divide to isolate the variable.
• Check your solution by plugging it back into the original equation.

Following these steps ensures a systematic approach, reducing the chance of error. In our exercise, noticing and correctly applying each of these steps led us to the solution \(x=0\). This process not only provides us with the answer but also reinforces our problem-solving skills for future equations.