Verified Answer :

  • Isolate \(x\) in \(x+3=7\) by subtracting 3 to get \(x = 4\).
  • Isolate \(x\) in \(\frac{x}{3}=7\) by multiplying by 3 to get \(x = 21\).
  • Isolate \(x\) in \(3x+1=13\) by subtracting 1 and dividing by 3 to get \(x = 4\).
  • Isolate \(x\) in \(10x-15=4x+27\) by gathering like terms to get \(x = 7\).
  • The final solutions are \(\boxed{4, 21, 4, 7}\).

Explanation

  1. Problem Analysis

The problem presents four distinct, independent linear equations in the variable \(x\). Our goal is to solve for \(x\) in each individual case. Each equation is of the form \(ax + b = cx + d\), which we can solve by isolating the variable \(x\) using standard algebraic manipulations, such as performing the same arithmetic operation on both sides of the equation.

  1. Solving the Equations
  2. Equation 1: \(x + 3 = 7\). To isolate \(x\), we subtract 3 from both sides:

\(x = 7 - 3 = 4\).

  1. Equation 2: \(\frac{x}{3} = 7\). To isolate \(x\), we multiply both sides by 3:

\(x = 7 \times 3 = 21\).

  1. Equation 3: \(3x + 1 = 13\). First, subtract 1 from both sides:

\(3x = 13 - 1 = 12\).
Then, divide both sides by 3:
\(x = \frac{12}{3} = 4\).

  1. Equation 4: \(10x - 15 = 4x + 27\). First, subtract \(4x\) from both sides to group the \(x\) terms:

\(10x - 4x - 15 = 27\), which simplifies to \(6x - 15 = 27\).
Next, add 15 to both sides to move the constant term:
\(6x = 27 + 15 = 42\).
Finally, divide by 6:
\(x = \frac{42}{6} = 7\).

  1. Final Results

By performing the required algebraic steps for each equation, we have determined the value of \(x\) for each case:

  • For \(x + 3 = 7\), \(x = 4\).
  • For \(\frac{x}{3} = 7\), \(x = 21\).
  • For \(3x + 1 = 13\), \(x = 4\).
  • For \(10x - 15 = 4x + 27\), \(x = 7\).

The solutions are \(\boxed{4, 21, 4, 7}\).

Examples

Solving linear equations is a fundamental skill used daily to manage resources or determine missing variables. For example, if you are saving money for a bicycle (x) and you currently have \(\$3\) saved and need \(\$7\) total, the equation \(x + 3 = 7\) models your situation. Similarly, if you want to distribute your total collection of marbles (\(x\)) into 3 groups and each group needs to have 7 marbles, \(\frac{x}{3} = 7\) helps you find your total collection size. Mastering these simple algebraic steps enables you to solve real-world problems involving balancing budgets, shopping, and planning.

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