1. Problem: Given a binary operation $*$ on $\mathbb{R}$ defined by $x * y = 2x + 2y - \frac{xy}{5}$, find: (a) The inverse of $x$ under $*$. (b) The truth set when $m * 7 = -2 * m$. 2. Problem: Given $\binom{x}{2} + \binom{x}{3} - 4x = 0$ where $x$ is a positive integer, find $x$. 3. Problem: Given $f(x) = \frac{72}{px + r}$ with constants $p, r$, and $f(6) = 12$, $f(7) = 9$, find: (a) The values of $p$ and $r$. (b) $f^{-1}\left(\frac{5}{3}\right)$. 4. Problem: Resolve $\frac{2x - 7}{25x - 24 - 6x^2}$ into partial fractions. --- 1. (a) Find inverse of $x$ under $*$: We want $x * x^{-1} = e$, where $e$ is the identity element. First, find $e$ such that $x * e = x$ for all $x$: $$x * e = 2x + 2e - \frac{xe}{5} = x$$ Rearranged: $$2x + 2e - \frac{xe}{5} = x \implies 2e - \frac{xe}{5} = x - 2x = -x$$ Since this must hold for all $x$, separate terms: $$2e - \frac{xe}{5} = -x$$ Rewrite as: $$2e = -x + \frac{xe}{5} = x\left(-1 + \frac{e}{5}\right)$$ For this to hold for all $x$, the coefficient of $x$ on the right must be zero, so: $$-1 + \frac{e}{5} = 0 \implies e = 5$$ Check $2e = 2 \times 5 = 10$ matches left side. So identity element is $e = 5$. Now find $x^{-1}$ such that: $$x * x^{-1} = 5$$ Using definition: $$2x + 2x^{-1} - \frac{x \cdot x^{-1}}{5} = 5$$ Let $y = x^{-1}$, then: $$2x + 2y - \frac{xy}{5} = 5$$ Rearranged: $$2y - \frac{xy}{5} = 5 - 2x$$ Factor $y$: $$y\left(2 - \frac{x}{5}\right) = 5 - 2x$$ Simplify inside parentheses: $$2 - \frac{x}{5} = \frac{10 - x}{5}$$ So: $$y \cdot \frac{10 - x}{5} = 5 - 2x$$ Multiply both sides by $\frac{5}{10 - x}$: $$y = \frac{5(5 - 2x)}{10 - x}$$ Thus, $$\boxed{x^{-1} = \frac{5(5 - 2x)}{10 - x}}$$ --- 1. (b) Find truth set for $m * 7 = -2 * m$: Calculate each side: $$m * 7 = 2m + 2 \times 7 - \frac{m \times 7}{5} = 2m + 14 - \frac{7m}{5}$$ $$-2 * m = 2(-2) + 2m - \frac{(-2) m}{5} = -4 + 2m + \frac{2m}{5}$$ Set equal: $$2m + 14 - \frac{7m}{5} = -4 + 2m + \frac{2m}{5}$$ Subtract $2m$ from both sides: $$14 - \frac{7m}{5} = -4 + \frac{2m}{5}$$ Bring constants to one side and $m$ terms to the other: $$14 + 4 = \frac{2m}{5} + \frac{7m}{5}$$ $$18 = \frac{9m}{5}$$ Multiply both sides by $\frac{5}{9}$: $$m = \frac{18 \times 5}{9} = 10$$ Truth set is $\boxed{\{10\}}$. --- 2. Solve $\binom{x}{2} + \binom{x}{3} - 4x = 0$ for positive integer $x$: Recall: $$\binom{x}{2} = \frac{x(x-1)}{2}, \quad \binom{x}{3} = \frac{x(x-1)(x-2)}{6}$$ Substitute: $$\frac{x(x-1)}{2} + \frac{x(x-1)(x-2)}{6} - 4x = 0$$ Multiply entire equation by 6 to clear denominators: $$3x(x-1) + x(x-1)(x-2) - 24x = 0$$ Expand terms: $$3x^2 - 3x + x^3 - 3x^2 + 2x - 24x = 0$$ Simplify: $$x^3 - 3x^2 + 3x^2 - 3x + 2x - 24x = 0 \implies x^3 - 25x = 0$$ Factor: $$x(x^2 - 25) = 0$$ So: $$x = 0 \quad \text{or} \quad x^2 = 25 \implies x = \pm 5$$ Since $x$ is positive integer, $x = 5$. Answer: $\boxed{5}$. --- 3. Given $f(x) = \frac{72}{px + r}$, with $f(6) = 12$ and $f(7) = 9$. (a) Find $p$ and $r$. Use given values: $$f(6) = \frac{72}{6p + r} = 12 \implies 6p + r = \frac{72}{12} = 6$$ $$f(7) = \frac{72}{7p + r} = 9 \implies 7p + r = \frac{72}{9} = 8$$ Subtract first from second: $$(7p + r) - (6p + r) = 8 - 6 \implies p = 2$$ Substitute $p=2$ into $6p + r = 6$: $$6 \times 2 + r = 6 \implies 12 + r = 6 \implies r = -6$$ Answer: $$\boxed{p = 2, r = -6}$$ (b) Find $f^{-1}\left(\frac{5}{3}\right)$. Set $f(x) = \frac{5}{3}$: $$\frac{72}{2x - 6} = \frac{5}{3}$$ Cross multiply: $$72 \times 3 = 5(2x - 6)$$ $$216 = 10x - 30$$ Add 30: $$246 = 10x$$ Divide by 10: $$x = 24.6$$ Answer: $$\boxed{24.6}$$ --- 4. Resolve $\frac{2x - 7}{25x - 24 - 6x^2}$ into partial fractions. Rewrite denominator: $$25x - 24 - 6x^2 = -6x^2 + 25x - 24$$ Multiply numerator and denominator by $-1$ to get standard form: $$\frac{2x - 7}{-6x^2 + 25x - 24} = \frac{-(2x - 7)}{6x^2 - 25x + 24} = \frac{-2x + 7}{6x^2 - 25x + 24}$$ Factor denominator: Find factors of $6 \times 24 = 144$ that sum to $-25$: $$-9 \text{ and } -16$$ Rewrite: $$6x^2 - 9x - 16x + 24 = 3x(2x - 3) - 8(2x - 3) = (3x - 8)(2x - 3)$$ So: $$\frac{-2x + 7}{(3x - 8)(2x - 3)} = \frac{A}{3x - 8} + \frac{B}{2x - 3}$$ Multiply both sides by denominator: $$-2x + 7 = A(2x - 3) + B(3x - 8)$$ Expand: $$-2x + 7 = 2Ax - 3A + 3Bx - 8B = (2A + 3B)x + (-3A - 8B)$$ Equate coefficients: $$-2 = 2A + 3B$$ $$7 = -3A - 8B$$ Solve system: From first: $$2A = -2 - 3B \implies A = \frac{-2 - 3B}{2}$$ Substitute into second: $$7 = -3 \times \frac{-2 - 3B}{2} - 8B = \frac{6 + 9B}{2} - 8B = \frac{6 + 9B - 16B}{2} = \frac{6 - 7B}{2}$$ Multiply both sides by 2: $$14 = 6 - 7B \implies 14 - 6 = -7B \implies 8 = -7B \implies B = -\frac{8}{7}$$ Find $A$: $$A = \frac{-2 - 3(-8/7)}{2} = \frac{-2 + 24/7}{2} = \frac{-\frac{14}{7} + \frac{24}{7}}{2} = \frac{10/7}{2} = \frac{10}{14} = \frac{5}{7}$$ Answer: $$\boxed{\frac{2x - 7}{25x - 24 - 6x^2} = \frac{5/7}{3x - 8} - \frac{8/7}{2x - 3}}$$