- The first equation \(\frac{x}{x} + 3 = 7\) simplifies to \(4 = 7\), which is a contradiction, meaning it has no solution.
- Solving \(\frac{x}{3} = 7\) leads to \(x = \boxed{21}\).
- Solving \(3x + 1 = 13\) leads to \(x = \boxed{4}\).
- Solving \(10x - 15 = 4x + 27\) leads to \(x = \boxed{7}\).
Explanation
- Analysis of the equations
The problem presents four independent equations involving the variable \(x\). We will solve each one separately:
- \(\frac{x}{x} + 3 = 7\)
- \(\frac{x}{3} = 7\)
- \(3x + 1 = 13\)
- \(10x - 15 = 4x + 27\)
We need to solve these one by one and observe the results.
- Solving the first equation
For the first equation \(\frac{x}{x} + 3 = 7\), we note that for any \(x
eq 0\), the term \(\frac{x}{x}\) simplifies to 1. Substituting this into the equation, we obtain:
\(1 + 3 = 7\)
\(4 = 7\)
Since \(4\) is not equal to \(7\), this equation is a contradiction and has no solution. The condition \(x
eq 0\) was required for the expression to be defined, but even then, the statement is false.
- Solving the second equation
For the second equation \(\frac{x}{3} = 7\), we multiply both sides by 3 to isolate \(x\):
\(x = 7 \times 3\)
\(x = 21\)
Thus, the solution for the second equation is \(x = 21\).
- Solving the third equation
For the third equation \(3x + 1 = 13\), we first subtract 1 from both sides:
\(3x = 13 - 1\)
\(3x = 12\)
Next, we divide both sides by 3:
\(x = \frac{12}{3}\)
\(x = 4\)
Thus, the solution for the third equation is \(x = 4\).
- Solving the fourth equation
For the fourth equation \(10x - 15 = 4x + 27\), we group the \(x\) terms on the left and the constants on the right:
\(10x - 4x = 27 + 15\)
\(6x = 42\)
Dividing both sides by 6 gives:
\(x = \frac{42}{6}\)
\(x = 7\)
Thus, the solution for the fourth equation is \(x = 7\). In summary, the first equation has no solution, and the solutions for the others are \(x=21\), \(x=4\), and \(x=7\), respectively.
Examples
These equations demonstrate the basics of isolating variables, which is a foundational skill in algebra. In real life, this is used in everything from calculating your remaining budget after expenses (like the \(3x + 1 = 13\) type) to scaling a recipe where if you multiply your ingredients, the resulting total must balance correctly. For example, if you know that one third of a recipe makes 7 servings, finding that the full recipe makes \(x=21\) servings helps you plan for large parties. Understanding how to handle these variations is key to logical problem-solving.