Key Concepts:
Linear Equations:
Equations of the form \(ax + b = 0\), where \(a \neq 0\).Functional Equations:
Equations involving functions where the goal is to determine the function form, e.g., \(f(x+1) = 2f(x)\).Inequalities:
Statements involving \(<\), \(>\), \(\leq\), or \(\geq\) that describe the range of solutions for a variable.Example 1: Mary has \\(3.00 more than Ben but \\)5.00 less than Jane. If Mary has \$x, how much do Jane and Ben have altogether?
Solution:
Step 1: Define the amounts Jane and Ben have in terms of Mary: \[ \text{Ben's amount: } x - 3, \quad \text{Jane's amount: } x + 5. \] Step 2: Add their amounts: \[ \text{Total: } (x - 3) + x + (x + 5). \] Step 3: Simplify: \[ \text{Total: } 3x + 2. \] Therefore, Jane and Ben have \$\(3x + 2\) altogether.Example 2: Solve the linear equation \(2x - 5 = 7\).
Solution:
Step 1: Add 5 to both sides: \[ 2x - 5 + 5 = 7 + 5 \implies 2x = 12. \] Step 2: Divide by 2: \[ x = \frac{12}{2} = 6. \] Therefore, the solution is \(x = 6\).Example 3: Solve the inequality \(2x - 3 > 7\).
Solution:
Step 1: Add 3 to both sides: \[ 2x - 3 + 3 > 7 + 3 \implies 2x > 10. \] Step 2: Divide by 2: \[ x > \frac{10}{2} \implies x > 5. \] Therefore, the solution is \(x > 5\).Example 4: Linear Equation with Irrational Coefficients
Solve: \[ \sqrt{50}x - \sqrt{2} = 0 \]Solution:
Step 1: Simplify \(\sqrt{50}\): \[ \sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2} \] Now the equation becomes: \[ 5\sqrt{2}x - \sqrt{2} = 0 \] Step 2: Add \(\sqrt{2}\) to both sides: \[ 5\sqrt{2}x = \sqrt{2} \] Step 3: Divide both sides by \(5\sqrt{2}\): \[ x = \frac{\sqrt{2}}{5\sqrt{2}} = \frac{1}{5} \] Therefore, the solution is \( \boxed{x = \frac{1}{5}} \).Example 5: Solve the functional equation \(f(x+1) = 2f(x)\), given \(f(1) = 3\).
Solution:
Step 1: Write the values of \(f(x)\) using the functional equation: \[ f(x+1) = 2f(x). \] Step 2: Find \(f(2)\) using \(f(1) = 3\): \[ f(2) = 2f(1) = 2 \cdot 3 = 6. \] Step 3: Find \(f(3)\) using \(f(2) = 6\): \[ f(3) = 2f(2) = 2 \cdot 6 = 12. \] Step 4: Generalize the pattern: \[ f(x+1) = 2f(x) \implies f(x) = 3 \cdot 2^{x-1}. \] Therefore, the solution to the functional equation is: \[ f(x) = 3 \cdot 2^{x-1}. \]