One Solution Condition 3154Eb

1. **State the problem:** We are given the equation $$-8 (x + 2) = h (-x - 2)$$ and need to find the value of $h$ such that the equation has exactly one solution. 2. **Rewrite the equation:** Distribute both sides: $$-8x - 16 = -hx - 2h$$ 3. **Group like terms:** Bring all $x$ terms to one side and constants to the other: $$-8x + hx = -2h + 16$$ 4. **Factor out $x$ on the left side:** $$x(-8 + h) = -2h + 16$$ 5. **Analyze the number of solutions:** - If the coefficient of $x$ is not zero, i.e., $$-8 + h \neq 0$$, then we can solve for $x$ uniquely: $$x = \frac{-2h + 16}{-8 + h}$$ - If the coefficient of $x$ is zero, i.e., $$-8 + h = 0 \Rightarrow h = 8$$, then the equation becomes: $$0 = -2(8) + 16 = -16 + 16 = 0$$ which is true for all $x$, so infinite solutions. - If the constant side is not zero when $h=8$, no solution. 6. **Conclusion:** - The equation has exactly one solution when $$h \neq 8$$ because then there is one $x$ term on each side and the equation can be solved uniquely. - When $$h = 8$$, the $x$ terms cancel out and the equation either has infinite solutions or no solution depending on constants. **Final answer:** This equation will have one solution when $h \neq 8$ because you get one solution when you have one number of $x$'s on either side of the equation and the coefficients of $x$ are not equal.