Day 1 - Exercise 1 Solutions
Solution to Question 1
Let’s solve the equation \(2x + 3 = 7\).
Subtract 3 from both sides:
\(2x + 3 - 3 = 7 - 3 \Rightarrow 2x = 4\)
Divide both sides by 2:
\(x = 2\)
Solution to Question 2
Given the equation \(x^2 - x - 6 = 0\), factor it.
We find factors of -6 that add up to -1. These are -3 and 2.
Therefore, \(x^2 - x - 6 = (x - 3)(x + 2)\)
Setting each factor to zero gives us the solutions: \(x = 3\) or \(x = -2\).
Solution to Question 3
The function \(f(x) = 3x + 4\) represents a straight line.
For \(x = 0\), \(f(0) = 4\) (the y-intercept).
For \(x = 1\), \(f(1) = 7\).
The line passes through points (0,4) and (1,7).