Jamiya Weber

Answered

2022-06-25

A fraction problem

$a=x+\frac{1}{x}\phantom{\rule{0ex}{0ex}}b=y+\frac{1}{y}\phantom{\rule{0ex}{0ex}}c=xy+\frac{1}{xy}$

Express $c$ in terms of $a$ and $b$

$a=x+\frac{1}{x}\phantom{\rule{0ex}{0ex}}b=y+\frac{1}{y}\phantom{\rule{0ex}{0ex}}c=xy+\frac{1}{xy}$

Express $c$ in terms of $a$ and $b$

Answer & Explanation

assumintdz

Expert

2022-06-26Added 22 answers

Note that $c$ is not uniquely specified by $a$ and $b$, since solving $a=x+\frac{1}{x}$ for $x$ yields two solutions which are reciprocals of each other, and applying $x\mapsto \frac{1}{x}$ or $y\mapsto \frac{1}{y}$ to $xy+\frac{1}{xy}$ yields $\frac{x}{y}+\frac{y}{x}$. However, applying either substitution again gives back $xy+\frac{1}{xy}$, so $c$ can take on either of those values.

Let ${c}_{1}=xy+\frac{1}{xy}$ and ${c}_{2}=\frac{x}{y}+\frac{y}{x}$. One can check that

${c}_{1}+{c}_{2}=ab$

and

${c}_{1}{c}_{2}={a}^{2}+{b}^{2}-4.$

Use the quadratic formula to solve for the possible values of $c$

Let ${c}_{1}=xy+\frac{1}{xy}$ and ${c}_{2}=\frac{x}{y}+\frac{y}{x}$. One can check that

${c}_{1}+{c}_{2}=ab$

and

${c}_{1}{c}_{2}={a}^{2}+{b}^{2}-4.$

Use the quadratic formula to solve for the possible values of $c$

Semaj Christian

Expert

2022-06-27Added 12 answers

Multiply all things by the LCD to remove fractions.

$ax={x}^{2}+1$

$by={y}^{2}+1$

$cxy={x}^{2}{y}^{2}+1$

Multiply the first two equations to get

$abxy={x}^{2}{y}^{2}+1+{x}^{2}+{y}^{2}=c+{x}^{2}+{y}^{2}$

Adding the first two equations, we get

$ax+by={x}^{2}+{y}^{2}+2$

Combining these two lines,

$abxy=c+ax+by-2$

At this point, I'd just go back and solve for $x,y$ using the quadratic formula,

$c+a\left(\frac{a\pm \sqrt{{a}^{2}-4}}{2}\right)+b\left(\frac{b\pm \sqrt{{b}^{2}-4}}{2}\right)-2=ab\left(\frac{a\pm \sqrt{{a}^{2}-4}}{2}\right)\left(\frac{b\pm \sqrt{{b}^{2}-4}}{2}\right)$

Not the most beautiful thing in the world, but at least you don't have to square the thing or anything messy like that.

$ax={x}^{2}+1$

$by={y}^{2}+1$

$cxy={x}^{2}{y}^{2}+1$

Multiply the first two equations to get

$abxy={x}^{2}{y}^{2}+1+{x}^{2}+{y}^{2}=c+{x}^{2}+{y}^{2}$

Adding the first two equations, we get

$ax+by={x}^{2}+{y}^{2}+2$

Combining these two lines,

$abxy=c+ax+by-2$

At this point, I'd just go back and solve for $x,y$ using the quadratic formula,

$c+a\left(\frac{a\pm \sqrt{{a}^{2}-4}}{2}\right)+b\left(\frac{b\pm \sqrt{{b}^{2}-4}}{2}\right)-2=ab\left(\frac{a\pm \sqrt{{a}^{2}-4}}{2}\right)\left(\frac{b\pm \sqrt{{b}^{2}-4}}{2}\right)$

Not the most beautiful thing in the world, but at least you don't have to square the thing or anything messy like that.

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