1. **State the problem:** We are given a system of linear equations: $$\begin{cases} 3g + 3a = 0 \\ 6g + 3k + 3b = 0 \\ 3g + 6k + 3l - 3a + 3c = 0 \\ 2g - 3k + 6l + 9 - 3b + 3d = 0 \\ 2g - 2k - 3l + 6 - 3c = 0 \\ 2k - 2l + c - 3d = 0 \\ 2l + d = 1 \end{cases}$$ 2. **Goal:** Solve for variables $g, a, k, b, l, c, d$. 3. **Step-by-step solution:** **Step 1:** From the first equation: $$3g + 3a = 0 \implies g + a = 0 \implies a = -g$$ **Step 2:** Substitute $a = -g$ into the third equation: $$3g + 6k + 3l - 3(-g) + 3c = 0 \implies 3g + 6k + 3l + 3g + 3c = 0 \implies 6g + 6k + 3l + 3c = 0$$ Divide by 3: $$2g + 2k + l + c = 0$$ **Step 3:** From the second equation: $$6g + 3k + 3b = 0 \implies 2g + k + b = 0 \implies b = -2g - k$$ **Step 4:** Substitute $b$ into the fourth equation: $$2g - 3k + 6l + 9 - 3b + 3d = 0$$ Replace $b$: $$2g - 3k + 6l + 9 - 3(-2g - k) + 3d = 0$$ Simplify: $$2g - 3k + 6l + 9 + 6g + 3k + 3d = 0$$ Combine like terms: $$8g + 6l + 9 + 3d = 0$$ **Step 5:** From the fifth equation: $$2g - 2k - 3l + 6 - 3c = 0$$ **Step 6:** From the sixth equation: $$2k - 2l + c - 3d = 0$$ **Step 7:** From the seventh equation: $$2l + d = 1 \implies d = 1 - 2l$$ **Step 8:** Substitute $d$ into the equation from Step 4: $$8g + 6l + 9 + 3(1 - 2l) = 0$$ Simplify: $$8g + 6l + 9 + 3 - 6l = 0 \implies 8g + 12 = 0 \implies 8g = -12 \implies g = -\frac{12}{8} = -\frac{3}{2}$$ **Step 9:** From Step 1, $a = -g = \frac{3}{2}$. **Step 10:** Substitute $g$ into the equation from Step 2: $$2(-\frac{3}{2}) + 2k + l + c = 0 \implies -3 + 2k + l + c = 0 \implies 2k + l + c = 3$$ **Step 11:** From Step 5: $$2g - 2k - 3l + 6 - 3c = 0$$ Substitute $g = -\frac{3}{2}$: $$2(-\frac{3}{2}) - 2k - 3l + 6 - 3c = 0 \implies -3 - 2k - 3l + 6 - 3c = 0$$ Simplify: $$3 - 2k - 3l - 3c = 0 \implies -2k - 3l - 3c = -3$$ **Step 12:** From Step 6: $$2k - 2l + c - 3d = 0$$ Substitute $d = 1 - 2l$: $$2k - 2l + c - 3(1 - 2l) = 0 \implies 2k - 2l + c - 3 + 6l = 0$$ Simplify: $$2k + 4l + c - 3 = 0 \implies 2k + 4l + c = 3$$ **Step 13:** Now we have three equations with variables $k, l, c$: $$\begin{cases} 2k + l + c = 3 \\ -2k - 3l - 3c = -3 \\ 2k + 4l + c = 3 \end{cases}$$ **Step 14:** Add the first and second equations: $$(2k + l + c) + (-2k - 3l - 3c) = 3 + (-3) \implies (0k) + (-2l) + (-2c) = 0 \implies -2l - 2c = 0 \implies l + c = 0 \implies c = -l$$ **Step 15:** Substitute $c = -l$ into the first and third equations: - First: $$2k + l + (-l) = 3 \implies 2k = 3 \implies k = \frac{3}{2}$$ - Third: $$2k + 4l + (-l) = 3 \implies 2k + 3l = 3$$ Substitute $k = \frac{3}{2}$: $$2(\frac{3}{2}) + 3l = 3 \implies 3 + 3l = 3 \implies 3l = 0 \implies l = 0$$ **Step 16:** From $c = -l$, $c = 0$. **Step 17:** From Step 7, $d = 1 - 2l = 1 - 0 = 1$. **Step 18:** From Step 3, $b = -2g - k = -2(-\frac{3}{2}) - \frac{3}{2} = 3 - \frac{3}{2} = \frac{3}{2}$. **Final solution:** $$g = -\frac{3}{2}, a = \frac{3}{2}, k = \frac{3}{2}, b = \frac{3}{2}, l = 0, c = 0, d = 1$$