1. **Stating the problem:** We are given a polynomial expression with variables $g, k, a, c, d, e, t, r$ and a system of equations relating these variables. The goal is to simplify the polynomial using the given relations and find the value of $r$. 2. **Given relations:** - $g = -a$ - $3g + 3a = 0$ - $6g + 3k + 36 = 0$ - $2g + 76 + 6 = 0$ - $3g + 6k + 3c - 3g + 3c = 0$ - $2g - 3k + 6e + a - 36 + 3d = 0$ - $2g - 7k - 3e + 6 - 3c = 0$ - $2k - 2e + c - 3d = 0$ - $2e + d = r$ - $d = r - 2e$ - $egin{cases} 2g + k + 6 = 0 \ g + 2k + t - a + c = 0 \ -2a + k + 6 = 0 \ -2a + 2k + e + c = 0 \ -k + 6 - e - c = 0 \ k = 6 - e + c \\ ext{and } g = -a \\ ext{from above} \\ ext{and } d = r - 2e \\ ext{and } 2e + d = r \\ ext{which is consistent} \end{cases}$ 3. **Step-by-step simplification:** - From $g = -a$, substitute $g$ in all equations. - From $3g + 3a = 0$, substituting $g = -a$ gives $3(-a) + 3a = 0$ which is true. - From $2g + k + 6 = 0$, substitute $g = -a$: $$2(-a) + k + 6 = 0 \Rightarrow -2a + k + 6 = 0 \Rightarrow k = 2a - 6$$ - From $-2a + k + 6 = 0$, same as above, consistent. - From $k = 6 - e + c$, and $k = 2a - 6$, equate: $$2a - 6 = 6 - e + c \Rightarrow 2a - 6 = 6 - e + c$$ Rearranged: $$2a - 6 - 6 + e - c = 0 \Rightarrow 2a - 12 + e - c = 0$$ - From $-k + 6 - e - c = 0$, substitute $k = 2a - 6$: $$-(2a - 6) + 6 - e - c = 0 \Rightarrow -2a + 6 + 6 - e - c = 0 \Rightarrow -2a + 12 - e - c = 0$$ - Adding the two equations: $$ (2a - 12 + e - c) + (-2a + 12 - e - c) = 0 + 0 \Rightarrow 0 - 0 - 2c = 0 \Rightarrow -2c = 0 \Rightarrow c = 0$$ - Substitute $c=0$ back into $2a - 12 + e - c = 0$: $$2a - 12 + e = 0 \Rightarrow e = 12 - 2a$$ - From $k = 6 - e + c$, with $c=0$: $$k = 6 - e = 6 - (12 - 2a) = 6 - 12 + 2a = 2a - 6$$ Consistent with previous $k$. - From $2k - 2e + c - 3d = 0$, substitute $k=2a-6$, $e=12-2a$, $c=0$: $$2(2a - 6) - 2(12 - 2a) + 0 - 3d = 0$$ $$4a - 12 - 24 + 4a - 3d = 0$$ $$8a - 36 - 3d = 0 \Rightarrow 3d = 8a - 36 \Rightarrow d = \frac{8a - 36}{3}$$ - From $2e + d = r$, substitute $e$ and $d$: $$2(12 - 2a) + \frac{8a - 36}{3} = r$$ $$24 - 4a + \frac{8a - 36}{3} = r$$ - Multiply all terms by 3 to clear denominator: $$3r = 3(24 - 4a) + 8a - 36 = 72 - 12a + 8a - 36 = 36 - 4a$$ - Therefore: $$r = \frac{36 - 4a}{3} = 12 - \frac{4a}{3}$$ 4. **Final answer:** $$\boxed{r = 12 - \frac{4a}{3}}$$ This expresses $r$ in terms of $a$ using the given system and substitutions.