Linear system with a parameter (elimination)

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Given the system with parameter $a$: $\begin{cases} 5x-4y=-2\\ 2x+ay=25 \end{cases}$

• (a) solve it after setting $a=7$;
• (b) show that for $a=-\dfrac{8}{5}$ the system is impossible.

Solution

(a) With $a=7$: $\begin{cases} 5x-4y=-2\\ 2x+7y=25 \end{cases}$. Using elimination, multiply the first by $2$ and the second by $5$: $\begin{cases} 10x-8y=-4\\ 10x+35y=125 \end{cases}$ Subtracting the first from the second: $43y=129 \implies y=3$. Substituting into the first: $5x-12=-2 \implies x=2$. $\boxed{x=2,\quad y=3}$

(b) With $a=-\dfrac85$ the second equation is $2x-\dfrac85 y=25$; multiplying it by $\dfrac52$ gives $5x-4y=\dfrac{125}{2}$. But the first equation requires $5x-4y=-2$. The same combination $5x-4y$ would have to equal both $-2$ and $\dfrac{125}{2}$: a contradiction. The system is therefore impossible (the two lines are parallel). $\boxed{a=-\tfrac85 \Rightarrow \text{impossible system}}$