Subjects algebra

Substitution Method 8B65E0

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1. **State the problem:** Solve the system of equations using the substitution method: $$\begin{cases}-5x - 8y = 17 \\ 2x - 7y = -17\end{cases}$$ 2. **Isolate one variable in one equation:** From the second equation, solve for $x$: $$2x - 7y = -17$$ Add $7y$ to both sides: $$2x = 7y - 17$$ Divide both sides by 2: $$x = \frac{7y - 17}{2}$$ 3. **Substitute into the first equation:** Replace $x$ in the first equation with the expression found: $$-5\left(\frac{7y - 17}{2}\right) - 8y = 17$$ 4. **Simplify and solve for $y$:** Multiply out: $$-\frac{5}{2}(7y - 17) - 8y = 17$$ Distribute: $$-\frac{35y}{2} + \frac{85}{2} - 8y = 17$$ Convert $-8y$ to fraction with denominator 2: $$-\frac{35y}{2} - \frac{16y}{2} + \frac{85}{2} = 17$$ Combine like terms: $$-\frac{51y}{2} + \frac{85}{2} = 17$$ Multiply both sides by 2 to clear denominators: $$\cancel{2} \left(-\frac{51y}{\cancel{2}} + \frac{85}{\cancel{2}}\right) = 2 \times 17$$ $$-51y + 85 = 34$$ Subtract 85 from both sides: $$-51y = 34 - 85$$ $$-51y = -51$$ Divide both sides by $-51$: $$\frac{-51y}{\cancel{-51}} = \frac{-51}{\cancel{-51}}$$ $$y = 1$$ 5. **Find $x$ using $y=1$:** Substitute back into $x = \frac{7y - 17}{2}$: $$x = \frac{7(1) - 17}{2} = \frac{7 - 17}{2} = \frac{-10}{2} = -5$$ 6. **Final solution:** $$\boxed{(x, y) = (-5, 1)}$$ This means the solution to the system is $x = -5$ and $y = 1$.