1. The problem states: Given the quadratic equation $$ax^2 + a = b$$ where $a$ and $b$ are integers, and one solution is $x=6$, find the other solution. 2. Rewrite the equation as $$ax^2 + a - b = 0$$ to identify coefficients: $$A = a, B = 0, C = a - b$$. 3. Since $x=6$ is a root, substitute into the equation: $$a(6)^2 + a - b = 0$$ $$36a + a - b = 0$$ $$37a = b$$ 4. The quadratic equation becomes: $$ax^2 + a - 37a = 0$$ $$a x^2 + a - 37a = 0$$ $$a x^2 - 36a = 0$$ 5. Divide both sides by $a$ (assuming $a \neq 0$): $$\cancel{a} x^2 - 36 \cancel{a} = 0$$ $$x^2 - 36 = 0$$ 6. Solve for $x$: $$x^2 = 36$$ $$x = \pm 6$$ 7. Since one solution is $6$, the other solution is $$-6$$. Final answer: The other solution is $-6$.