Problem

Evaluate 3(x-6)=100+x

The problem given is an algebraic equation that requires you to find the value of the variable x. The equation sets the expression 3 times the quantity (x minus 6) equal to 100 plus x. To solve the equation, you would typically perform algebraic operations such as distributing the multiplication across the parentheses, isolating the variable on one side of the equation, and simplifying both sides to find the value of x that makes the equation true.

$3 \left(\right. x - 6 \left.\right) = 100 + x$

Answer

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Solution:

Step 1: Expand the expression on the left side of the equation.

  • Step 1.1: Start with the original equation. $0 + 0 + 3(x - 6) = 100 + x$
  • Step 1.2: Recognize that adding zero does not change the value. $3(x - 6) = 100 + x$
  • Step 1.3: Distribute the $3$ across the parentheses. $3x - 3 \cdot 6 = 100 + x$
  • Step 1.4: Perform the multiplication. $3x - 18 = 100 + x$

Step 2: Isolate the variable terms on one side.

  • Step 2.1: Subtract $x$ from both sides to start grouping $x$ terms. $3x - x - 18 = 100$
  • Step 2.2: Combine like terms. $2x - 18 = 100$

Step 3: Move the constant term to the opposite side.

  • Step 3.1: Add $18$ to both sides to isolate the variable term. $2x = 100 + 18$
  • Step 3.2: Combine the constants on the right side. $2x = 118$

Step 4: Solve for the variable.

  • Step 4.1: Divide the equation by the coefficient of $x$ to solve for $x$. $\frac{2x}{2} = \frac{118}{2}$

  • Step 4.2: Simplify the equation by canceling out the $2$s on the left side.

    • Step 4.2.1: Cancel the $2$s. $\frac{\cancel{2}x}{\cancel{2}} = \frac{118}{2}$
    • Step 4.2.1.1: Simplify the left side. $x = \frac{118}{2}$
  • Step 4.3: Simplify the right side by dividing. $x = 59$

Knowledge Notes:

The problem-solving process involves solving a linear equation in one variable. The key steps are:

  1. Distributive Property: This property states that $a(b + c) = ab + ac$. It's used to expand expressions like $3(x - 6)$.

  2. Combining Like Terms: This involves simplifying expressions by adding or subtracting terms that have the same variable to the same power. For instance, $3x - x$ simplifies to $2x$.

  3. Isolating the Variable: To solve for $x$, we need to get $x$ by itself on one side of the equation. This often involves moving terms from one side of the equation to the other by performing inverse operations (e.g., if there's a $+x$ on one side, we subtract $x$ from both sides).

  4. Inverse Operations: These are operations that reverse the effect of another operation. For example, addition is the inverse of subtraction, and multiplication is the inverse of division.

  5. Solving Basic Algebraic Equations: The goal is to isolate the variable on one side to find its value. This usually involves a series of steps including distributing, combining like terms, and using inverse operations.

  6. Checking Solutions: After finding the value of $x$, you can check your solution by substituting it back into the original equation to see if it makes a true statement.

In this particular problem, we applied these principles to solve for $x$ in the equation $3(x - 6) = 100 + x$.

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