Dealing more efficiently with fractional forms in system of equations

Question

As an example, suppose we have to solve the following system of two equations and two unknowns:

$$ \begin{cases} -\frac{10}{x}-\frac{8}{y} &= \frac{8}{3} \\ -\frac{6}{x}+\frac{6}{y} &= -\frac{1}{3} \end{cases} $$

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My approach and solution

I opted to solve it by combination, referring to the first equation as (1) and the second equation as (2), I started by eliminating the y's: \begin{align} \frac{6}{8}(1)+(2): -\frac{60}{8x}-\frac{6}{x} &= \frac{6}{3}-\frac{1}{3} \\ \frac{-60-48}{8x} &= \frac{5}{3} \tag{*}\\ 40x &=3(-108) \tag{**} \\ x &=-\frac{81}{10} \tag{1'} \end{align} Substituting (1') into (2): \begin{align} \frac{100}{81}-\frac{8}{y}&=\frac{8}{3} \\ y &= -\frac{162}{29} \end{align} So I find the tuple of $(-\frac{81}{10};-\frac{162}{29})$ as solution.

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Questions

1. For such systems, should we not be concerned about the conditions of existence of the system? Namely, in this case both $x\neq 0$ and $y\neq 0,$ in the same way that we would do when solving an equation. If yes, how do we formally write the domain of existence for a system?
2. Is my transition from step $(*)$ to $(**)$ allowed? My understanding is that, yes for all $x\neq 0.$
3. I am really eager to learn whether there are simple ideas that simplify the system (and other similar systems) before we start solving it. As shown in my approach, it was slightly awkward dealing with the fractions throughout and the "large" numbers, which inherently may render the approach more prone to mistakes. Any alternative, quicker approach (I only know by combination and substitution) would be much appreciated.

Answer 1

As requested in a comment, here are the details of what I suggested in a comment. I've arranged the work to minimize the arithmetic details. In fact, I not only didn't have to resort to a calculator, but I did not have to resort to any "paper and pencil" multiplication. For example, $81 \cdot 2$ comes up at one point, but this can be done by sight: twice $80 + 1$ is $160 + 2.$

$$\begin{array}{r} -\frac{10}{x} \; - \; \frac{8}{y} & = & \frac{8}{3} \\ -\frac{6}{x} \; + \; \frac{6}{y} & = & -\frac{1}{3} \end{array}$$

Changing variables by letting $a = \frac{1}{x}$ and $b = \frac{1}{y}$ converts this into two linear equations in two unknowns.

$$\begin{array}{r} -10a \; - \; 8b & = & \frac{8}{3} \\ -6a \; + \; 6b & = & -\frac{1}{3} \end{array}$$

Now clear fractions by multiplying both sides of each equation by $3.$

$$\begin{array}{r} -30a \; - \; 24b & = & 8 \\ -18a \; + \; 18b & = & -1 \end{array}$$

Divide both sides of the first equation by $2.$

$$\begin{array}{r} -15a \; - \; 12b & = & 4 \\ -18a \; + \; 18b & = & -1 \end{array}$$

Ignoring signs, the coefficients of $a$ are $15 = 3 \cdot 5$ and $18 = 3 \cdot 6.$

Ignoring signs, the coefficients of $b$ are $12 = 6 \cdot 2$ and $18 = 6 \cdot 3.$

From this we can see that it will be simpler, when using the addition and subtraction method, to eliminate $b$ (multiply equations by $2$ and $3)$ than to eliminate $a$ (multiply equations by $5$ and $6).$

In the most recent pair of displayed equations, multiply both sides of the upper equation by $3$ and multiply both sides of the lower equation by $2.$

$$\begin{array}{r} -45a \; - \; 36b & = & 12 \\ -36a \; + \; 36b & = & -2 \end{array}$$

Adding the two equations above gives

$$ -81a \; = \; 10 \;\;\; \implies \;\;\; a = -\frac{10}{81} $$

Plugging this value of $a$ into what appears to be the simplest to-work-with equation above gives

$$ -18\left(-\frac{10}{81}\right) \; + \; 18b \; = \; -1 $$

$$ -2\left(-\frac{10}{9}\right) \; + \; 18b \; = \; -1 $$

Now clear fractions by multiplying both sides by $9,$ and continue in some standard way until $b$ is isolated.

$$ -2(-10) \; + \; 9(18b) \; = \; -9 $$

$$ 20 \; + \; (9 \cdot 9)(2b) \; = \; -9 $$

$$ 20 \; + \; 81(2b) \; = \; -9 $$

$$ 20 \; + \; 162b \; = \; -9 $$

$$ 162b \; = \; -9 - 20 \; = \; -29 $$

$$ b \; = \; -\frac{29}{162} $$

Therefore, we have $a = -\frac{10}{81} = \frac{1}{x}$ and $b = -\frac{29}{162} = \frac{1}{y},$ which implies $x = -\frac{81}{10}$ and $y = -\frac{162}{29}.$

Answer 2

$$ \begin{cases} -\frac{10}{x}-\frac{8}{y} &= \frac{8}{3},\quad{|}\cdot{3}\\ -\frac{6}{x}+\frac{6}{y} &= -\frac{1}{3}.\quad{|}\cdot{(-5)} \end{cases} \iff \begin{cases} -\frac{30}{x}-\frac{24}{y} &= \frac{24}{3},\\ \frac{30}{x}-\frac{30}{y} &= \frac{5}{3}. \end{cases} \iff $$ $$ \iff \begin{cases} -\frac{54}{y} &= \frac{29}{3},\\ \frac{30}{x}-\frac{30}{y} &= \frac{5}{3}. \end{cases} \iff \begin{cases} {y} = \frac{-162}{29},\\ \frac{30}{x}-\frac{30}{y} &= \frac{5}{3}. \end{cases} \iff \begin{cases} {y}=\frac{-162}{29},\\ {x}=\frac{-81}{10}. \end{cases} $$ Also the parameters $x\ne{0}$, $y\ne{0}$.

Good luck!

Answer 3

We start by getting rid of the fractions.

\begin{align*} \begin{cases} -\frac{10}{x}-\frac{8}{y} &= \frac{8}{3} &\implies -30y -24x=8xy \\ -\frac{6}{x}+\frac{6}{y} &= -\frac{1}{3} &\implies -18y+18x=-xy \end{cases} \\\text{We mulitply the second equation by 8 and then add} \\ \begin{cases} -\space \space 30y -\space \space 24x &=\space \space\space 8xy \\ -144y+144x &=-8xy &\implies 120 x - 174 y=0\\ &&\implies y=\dfrac{20x}{29} \end{cases} \end{align*}

By substitution \begin{align*} -30\cdot \dfrac{20x}{29} -24x&=8x\cdot \dfrac{20x}{29}\\ \implies -1296 x &= 160 x^2 &\implies x=\frac{-81}{10}\space \space \\ \\ -30y -24\cdot \frac{-81}{10} &= 8y\cdot \frac{-81}{10} \\ \implies -174y&=972 &\implies y= \dfrac{-162}{29}\\ \end{align*}

We confirm the calculations with WA here and here