1. **Stating the problem:** Solve the first system of equations simultaneously: $$\begin{cases} e^{2x} - 7e^{x+1} = -10 \\ x - y = 1 \end{cases}$$ 2. **Rewrite the first equation:** Recall that $e^{2x} = (e^x)^2$ and $e^{x+1} = e^x \cdot e^1 = e \cdot e^x$. So the first equation becomes: $$ (e^x)^2 - 7e \cdot e^x = -10 $$ 3. **Substitute:** Let $t = e^x$, then the equation is: $$ t^2 - 7e t = -10 $$ Rewrite as: $$ t^2 - 7e t + 10 = 0 $$ 4. **Solve quadratic in $t$:** Using the quadratic formula: $$ t = \frac{7e \pm \sqrt{(7e)^2 - 4 \cdot 1 \cdot 10}}{2} = \frac{7e \pm \sqrt{49e^2 - 40}}{2} $$ 5. **Calculate discriminant:** $$ \Delta = 49e^2 - 40 $$ Since $e \approx 2.718$, $49e^2$ is positive and larger than 40, so two real roots exist. 6. **Find $x$ from $t$:** Recall $t = e^x$, so: $$ x = \ln(t) $$ 7. **Use the second equation:** $$ x - y = 1 \implies y = x - 1 $$ 8. **Summary:** The solutions are: $$ x = \ln\left( \frac{7e \pm \sqrt{49e^2 - 40}}{2} \right), \quad y = x - 1 $$ --- **Now solve for $x$ in the second problem:** $$ e^{x+1} \cdot e^{x-3} = 2 $$ 1. **Combine exponents:** $$ e^{x+1} \cdot e^{x-3} = e^{(x+1)+(x-3)} = e^{2x - 2} $$ 2. **Set equal to 2:** $$ e^{2x - 2} = 2 $$ 3. **Take natural logarithm:** $$ 2x - 2 = \ln(2) $$ 4. **Solve for $x$:** $$ 2x = \ln(2) + 2 $$ $$ x = \frac{\ln(2) + 2}{2} $$ --- **Solve the third problem:** $$ h x + h y = h(x - 1) + h(y - 1) $$ 1. **Distribute $h$:** $$ h x + h y = h x - h + h y - h $$ 2. **Simplify:** $$ h x + h y = h x + h y - 2h $$ 3. **Subtract $h x + h y$ from both sides:** $$ 0 = -2h $$ 4. **This implies:** $$ 2h = 0 \implies h = 0 $$ --- **Final answers:** - For the first system: $$ x = \ln\left( \frac{7e \pm \sqrt{49e^2 - 40}}{2} \right), \quad y = x - 1 $$ - For the second equation: $$ x = \frac{\ln(2) + 2}{2} $$ - For the third equation: $$ h = 0 $$