1. Solve for \( x \) if: \( \begin{array}{ll}1.1 .1 & (2 x-1)(x+5)=0 \\ 1.1 .2 & x(3 x+13)=11 \\ 1.1 .3 & 1+\sqrt{2 x+1}=x \\ 1.1 .4 & 2^{x \sqrt{x}}=2^{27} \\ 1.1 .5 & x^{2}<-2 x+15 \\ \text { Solve for } x & \text { and } y \text { simultaneously if: } \\ 3-y+2 x=0 \text { and } 6 x+4 y^{2}=3+5 x y\end{array} \)

**1.1.1. Solve \( (2x-1)(x+5)=0 \)** Since the product is zero, set each factor equal to zero: \[ 2x-1 = 0 \quad \text{or} \quad x+5=0. \] Solve each: - For \(2x-1=0\): \[ 2x=1 \quad \Rightarrow \quad x=\frac{1}{2}. \] - For \(x+5=0\): \[ x=-5. \] Thus, the solutions are \( x=\frac{1}{2} \) and \( x=-5 \). --- **1.1.2. Solve \( x(3x+13)=11 \)** Expand the left‐hand side: \[ 3x^2 + 13x = 11. \] Bring all terms to one side: \[ 3x^2 + 13x - 11 = 0. \] Use the quadratic formula: \[ x=\frac{-13 \pm \sqrt{13^2-4\cdot3\cdot(-11)}}{2\cdot 3}. \] Compute the discriminant: \[ 13^2 -4\cdot3\cdot(-11)=169+132=301. \] Thus, \[ x=\frac{-13 \pm \sqrt{301}}{6}. \] --- **1.1.3. Solve \( 1+\sqrt{2x+1}=x \)** Isolate the square root: \[ \sqrt{2x+1} = x-1. \] **Note:** We require \( x-1\ge0 \) (i.e., \( x\ge1 \)) and also \( 2x+1\ge0 \) (which holds for \( x\ge -\frac{1}{2} \)). Thus, the primary restriction is \( x\ge1 \). Square both sides: \[ 2x+1 = (x-1)^2. \] Expand the right-hand side: \[ 2x+1 = x^2 -2x+1. \] Rearrange to form a quadratic: \[ 0 = x^2 -2x -2x +1 -1 = x^2 -4x. \] Factor: \[ x(x-4)=0. \] Thus, \( x=0 \) or \( x=4 \). However, \( x=0 \) does not satisfy \( x\ge1 \). Therefore, the only valid solution is \( x=4 \). --- **1.1.4. Solve \( 2^{x\sqrt{x}}=2^{27} \)** Since the bases are the same, we equate the exponents: \[ x\sqrt{x}=27. \] Express \( \sqrt{x} \) as \( x^{1/2} \): \[ x \cdot x^{1/2}= x^{3/2} = 27. \] Solve for \( x \) by raising both sides to the power \( \frac{2}{3} \): \[ x=27^{\frac{2}{3}}. \] Recognize that \( 27^{\frac{1}{3}}=3 \), thus: \[ x=3^2=9. \] --- **1.1.5. Solve the inequality \( x^2 < -2x+15 \)** First, bring all terms to one side: \[ x^2 + 2x -15 < 0. \] Find the roots of the quadratic equation \( x^2+2x-15=0 \) using the quadratic formula: \[ x=\frac{-2 \pm \sqrt{2^2-4\cdot1\cdot(-15)}}{2\cdot1}=\frac{-2 \pm \sqrt{4+60}}{2}=\frac{-2 \pm \sqrt{64}}{2}=\frac{-2 \pm 8}{2}. \] Thus, the roots are: \[ x = \frac{-2+8}{2}=3 \quad \text{and} \quad x = \frac{-2-8}{2}=-5. \] Since the quadratic coefficient is positive, the inequality holds between the roots: \[ -5 < x < 3. \] --- **Simultaneous Equations** Solve for \( x \) and \( y \) given: \[ \begin{cases} 3-y+2x=0,\\[1mm] 6x+4y^2=3+5xy. \end{cases} \] **Step 1: Express \( y \) in terms of \( x \) from the first equation.** \[ 3-y+2x=0 \quad \Rightarrow \quad y=3+2x. \] **Step 2: Substitute \( y=3+2x \) into the second equation.** \[ 6x + 4(3+2x)^2 = 3 + 5x(3+2x). \] Calculate \( (3+2x)^2 \): \[ (3+2x)^2=9+12x+4x^2. \] Thus, the equation becomes: \[ 6x+4(9+12x+4x^2)= 3+15x+10x^2. \] Simplify the left-hand side: \[ 6x + 36 + 48x + 16x^2 = 16x^2 + 54x + 36. \] So, \[ 16x^2 + 54x + 36 = 10x^2 + 15x+3. \] **Step 3: Bring all terms to one side.** \[ 16x^2 - 10x^2 + 54x - 15x + 36 - 3 = 0, \] \[ 6x^2 + 39x + 33 = 0. \] Divide the entire equation by 3: \[ 2x^2 + 13x + 11 = 0. \] **Step 4: Solve the quadratic equation using the quadratic formula.** \[ x=\frac{-13 \pm \sqrt{13^2-4\cdot2\cdot11}}{2\cdot2}=\frac{-13 \pm \sqrt{169-88}}{4}=\frac{-13 \pm \sqrt{81}}{4}=\frac{-13 \pm 9}{4}. \] Thus, the solutions for \( x \) are: - \( x=\frac{-13+9}{4}=\frac{-4}{4}=-1 \), - \( x=\frac{-13-9}{4}=\frac{-22}{4}=-\frac{11}{2} \). **Step 5: Find \( y \) using \( y= 3+2x \).** For \( x=-1 \): \[ y=3+2(-1)=3-2=1. \] For \( x=-\frac{11}{2} \): \[ y=3+2\left(-\frac{11}{2}\right)=3-11=-8. \] Thus, the solutions are: \[ (x,y)=(-1,\,1) \quad \text{and} \quad \left(-\frac{11}{2},\,-8\right). \]