QUESTION 1 DBE NOV 17 Q1 1.1 Solve for \( x \) : 1.1.1 \( \quad(2 x-3)(x+7)=0 \) 1.1.2 \( 7 x^{2}+3 x-2=0 \) (leave your answer correct to TWO decimal places) (3) 1.1.3 \( \sqrt{x-1}+3=x \) 1.1.4 \( \quad x^{2}>3(x+6) \) 1.2 Solve for \( x \) and \( y \) simultaneously: \[ \begin{array}{l} 2 y+x=1 \\ x^{2}+y^{2}+3 x y+y=0 \end{array} \]

To solve the quadratic equations and inequalities, it's crucial to remember that each step builds upon the last. For the equation \((2x - 3)(x + 7) = 0\), we can set each factor to zero, giving two potential solutions: \(x = \frac{3}{2}\) and \(x = -7\). For the quadratic equation \(7x^2 + 3x - 2 = 0\), apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find solutions approximately equal to \(x \approx 0.19\) and \(x \approx -1.57\). When solving \( \sqrt{x-1} + 3 = x \), isolate \( \sqrt{x-1} = x - 3 \) and then square both sides. This will yield the solution \(x \approx 4\). For the inequality \(x^2 > 3(x + 6)\), first rearrange to \(x^2 - 3x - 18 > 0\). This gives roots at \(x = 6\) and \(x = -3\). The solution sets can be expressed in intervals. Simultaneously solving the equations involves substitution or elimination. From \(2y + x = 1\), you can express \(x\) in terms of \(y\) (i.e., \(x = 1 - 2y\)) and substitute into the second equation, yielding a solvable quadratic that results in values for \(x\) and \(y\). --- For a bit of math nostalgia, did you know that the quadratic formula has its roots (no pun intended) back in ancient civilizations? The Babylonians were solving quadratic equations as far back as 2000 BC, using geometric methods to find solutions without the algebraic symbols we use today! If you're looking to improve your problem-solving skills, remember: always check your solutions by plugging them back into the original equations. It's easy to make mistakes when manipulating equations, especially with squaring both sides or handling negative numbers, so double-checking helps catch those pesky errors!