Solve the simultaneous equations: a^2 - ab + b^2 = 3 a + 2b + 1 = 0 Find the set of all real values of $x$ for which \[ \frac{2x - 5}{x - 3} \leq \frac{5}{2} \] - Leaving Cert Mathematics - Question 1 - 2012

From the second equation, we can express $a$ in terms of $b$:

$a = -2b - 1$

Substitute this value of $a$ into the first equation:

$(-2b - 1)^2 - (-2b - 1)b + b^2 = 3$

Expanding this expression gives:

$4b^2 + 4b + 1 + 2b^2 + b + b^2 = 3$

$7b^2 + 5b + 1 - 3 = 0$

$7b^2 + 5b - 2 = 0$

Using the quadratic formula where $b = 5$, $a = 7$, and $c = -2$:

$b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 7 \cdot (-2)}}{2 \cdot 7}$

$= \frac{-5 \pm \sqrt{25 + 56}}{14} = \frac{-5 \pm \sqrt{81}}{14}$

$= \frac{-5 \pm 9}{14}$

Thus, we have two potential solutions for $b$:

• $b = \frac{4}{14} = \frac{2}{7}$
• $b = \frac{-14}{14} = -1$

Calculate $a$ for these values of $b$:

• If $b = \frac{2}{7}$, then
  $a = -2\left(\frac{2}{7}\right) - 1 = \frac{-4}{7} - 1 = \frac{-4}{7} - \frac{7}{7} = \frac{-11}{7}$
• If $b = -1$, then $a = -2(-1) - 1 = 2 - 1 = 1$

Therefore, the solutions are:

$\{b = \frac{2}{7}, a = \frac{-11}{7}\} \text{ or } \{b = -1, a = 1\}$