Simultaneous Equations 81Adbc

1. **State the problem:** We need to solve the system of simultaneous equations: $$2x + y = 7$$ $$x - y = 1$$ Our goal is to find the values of $x$ and $y$ that satisfy both equations at the same time. 2. **Think first:** What happens if we add these two equations together? Let's pause and think about the $y$ terms. 3. **Add the equations:** $$\begin{aligned} &2x + y = 7 \\ +&x - y = 1 \\ \hline &3x + \cancel{y - y} = 8 \end{aligned}$$ Notice that $+y$ and $-y$ cancel each other out, leaving us with only $3x = 8$. 4. **Solve for $x$:** Divide both sides by 3: $$\frac{3x}{\cancel{3}} = \frac{8}{3} \implies x = \frac{8}{3}$$ 5. **Substitute $x$ back to find $y$:** Use the first equation: $$2\left(\frac{8}{3}\right) + y = 7$$ Simplify: $$\frac{16}{3} + y = 7$$ Subtract $\frac{16}{3}$ from both sides: $$y = 7 - \frac{16}{3} = \frac{21}{3} - \frac{16}{3} = \frac{5}{3}$$ 6. **Common mistake to watch out for:** Sometimes people incorrectly add $y + (-y)$ as $2y$. But remember, $y - y = 0$ because they cancel out. 7. **Verify the solution:** Check in the first equation: $$2\left(\frac{8}{3}\right) + \frac{5}{3} = \frac{16}{3} + \frac{5}{3} = \frac{21}{3} = 7$$ Check in the second equation: $$\frac{8}{3} - \frac{5}{3} = \frac{3}{3} = 1$$ Both equations are satisfied. 8. **Final answer:** $$x = \frac{8}{3}, \quad y = \frac{5}{3}$$ 9. **Quick recap:** 1. Write the equations clearly. 2. Add the equations to eliminate $y$. 3. Solve for $x$. 4. Substitute $x$ back to find $y$. 5. Check the answers. And that's it! Two equations, two unknowns, solved.