Dada la recta \( k x-y=k+3 \), determine un valor de \( k \) para que el punto \( (3,7) \) pertenezca dicha recta

7. \( 2 x^{2}+x-4=0 \) \( \begin{array}{ll}\text { a. }-\frac{1}{2} \pm \frac{\sqrt{33}}{4} & \text { c. }-\frac{1}{4} \pm \frac{\sqrt{33}}{4} \\ \text { b. }-4 \pm \frac{\sqrt{66}}{4} & \text { d. }-\frac{1}{2} \pm \frac{\sqrt{33}}{2}\end{array} \)

d) \( 16 \cdot 4^{15}:\left(8^{2} \cdot 4^{5}\right)=8^{x} \)

Solve the inequality. Express the solution set in set-builder notation and interval notation. Graph the solution set on a real number line. \[ 5(x-6)>4(x-6)+x \]

Solve the inequality. Express the solution set in set-builder notation and interval notation. Graph the solution set on a real number line. \( \frac{1}{2}(n-4)>n+5 \)

3. (*) Dado los vectores \( \vec{a}=\left(1,2 x_{1}, 3\right) \), \( \vec{b}=\left(-1,3 x_{2},-2 x_{1}\right), \vec{c}=\left(2,2,-3 x_{2}\right) \) los valores de \( x_{1} \quad y x_{2} \) para los cuales se cumple que \( 2 \vec{a}-3 \vec{b}+\vec{c}=(7,15,21) \) es respectivamente: a. \( (2,-1) \) b. \( (-1,2) \) c. \( (2,-3) \) d. \( (-3,1) \)