1.1.3 \( \quad 3\left(2 x^{2}-5\right)=x \) 1.2 Solve for \( x \) and \( y \) simultaneously: \( 4 x+y=-5 \) and \( -3 x+4 y=18 \)

### Problem 1.1.3 Solve the equation: \[ 3\left(2x^2 - 5\right)= x \] 1. **Distribute and simplify:** \[ 3 \cdot 2x^2 - 3 \cdot 5 = 6x^2 - 15 \] The equation becomes: \[ 6x^2 - 15 = x \] 2. **Bring all terms to one side:** \[ 6x^2 - x - 15 = 0 \] 3. **Apply the quadratic formula:** For a quadratic equation \(\,ax^2 + bx + c = 0\), the solutions are given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this equation, \(a = 6\), \(b = -1\), and \(c = -15\). 4. **Calculate the discriminant:** \[ \Delta = b^2 - 4ac = (-1)^2 - 4(6)(-15) = 1 + 360 = 361 \] 5. **Find the roots:** \[ x = \frac{-(-1) \pm \sqrt{361}}{2 \cdot 6} = \frac{1 \pm 19}{12} \] 6. **Compute the two solutions:** - For the positive root: \[ x = \frac{1 + 19}{12} = \frac{20}{12} = \frac{5}{3} \] - For the negative root: \[ x = \frac{1 - 19}{12} = \frac{-18}{12} = -\frac{3}{2} \] Thus, the solutions are: \[ x = \frac{5}{3} \quad \text{or} \quad x = -\frac{3}{2} \] --- ### Problem 1.2 Solve for \(x\) and \(y\) simultaneously: \[ 4x + y = -5 \quad \text{and} \quad -3x + 4y = 18 \] 1. **Express \(y\) in terms of \(x\) from the first equation:** \[ y = -5 - 4x \] 2. **Substitute \(y\) into the second equation:** \[ -3x + 4(-5 - 4x) = 18 \] Simplify: \[ -3x - 20 - 16x = 18 \] \[ -19x - 20 = 18 \] 3. **Solve for \(x\):** \[ -19x = 18 + 20 = 38 \] \[ x = \frac{38}{-19} = -2 \] 4. **Substitute \(x = -2\) back into \(y = -5 - 4x\):** \[ y = -5 - 4(-2) = -5 + 8 = 3 \] Thus, the solution is: \[ x = -2, \quad y = 3 \]