Solve the system 1/x + 1/y = 8 and 6/x + 12/y = 1.

Step-by-step solution

1. 1

   Substitute to remove the denominators. The unknowns only ever appear as $1/x$ and $1/y$, so setting $a=\frac1x,\qquad b=\frac1y$ converts the system into the linear one $a+b=8,\qquad 6a+12b=1.$ Clearing denominators by multiplying out would instead produce a messy quadratic system - the substitution is much cleaner.

2. 2

   Simplify the second equation. Dividing $6a+12b=1$ through by $6$ gives $a+2b=\frac16.$ Both equations now have the same coefficient on $a$, which sets up elimination in one subtraction.

3. 3

   Eliminate $a$ and solve for $b$. Subtracting the first equation from the second, $\left(a+2b\right)-\left(a+b\right)=\frac16-8\quad\Longrightarrow\quad b=\frac16-\frac{48}{6}=-\frac{47}{6}.$ A negative $b$ is perfectly acceptable - it just means $y$ will come out negative.

4. 4

   Back-substitute for $a$. From $a+b=8$, $a=8+\frac{47}{6}=\frac{48}{6}+\frac{47}{6}=\frac{95}{6}.$

5. 5

   Invert to recover $x$ and $y$. Since $a=1/x$ and $b=1/y$, $x=\frac{1}{a}=\frac{6}{95},\qquad y=\frac{1}{b}=-\frac{6}{47}.$ Both are nonzero, so the original denominators are legal and no solution has to be discarded.

6. 6

   Check in the original equations. $\dfrac{1}{x}=\dfrac{95}{6}$ and $\dfrac{1}{y}=-\dfrac{47}{6}$, so the first equation gives $\dfrac{95}{6}-\dfrac{47}{6}=\dfrac{48}{6}=8$, and the second gives $6\cdot\dfrac{95}{6}+12\cdot\left(-\dfrac{47}{6}\right)=95-94=1$. Both hold exactly, with no rounding involved.