Solving rational equations

Example 1: Solve the equation, $\dfrac{2}{x}-\dfrac{1}{3}=5$

Solution: First, find the LCD of all the terms in the equation. The LCD is $3x$.

Now multiply the equation by the LCD,

$3x\left(\dfrac{2}{x}-\dfrac{1}{3}\right)=3x\cdot5$

Simplifying,

$6-x=15x$

Solving for $x$,

$x=\dfrac{3}{8}$

Example 2: Solve, $\dfrac{4}{y-3}=2-\dfrac{7}{5y-15}$

Solution:

First, factor the denominators completely. We can factor $5y-15$ as $5y-15=5(y-3)$. So, we have

$\dfrac{4}{y-3}=2-\dfrac{7}{5(y-3)}$

Now find the LCD of all the terms in the equation. The LCD is $5(y-3)$.

Multiply by the LCD:

$20=10(y-3)-7$

or

$20=10y-30-7$

Solving,

$y=\dfrac{57}{10}$

Example 3: Solve $\dfrac{15}{x}=x+2$.

Solution: The LCD of the equation is $x$.

Multiplying the equation by the LCD,

$15=x^2+2x$

Rewriting the equation with zero on one side,

$x^2+2x-15=0$

This is a quadratic equation. Solving it, we get

$x=-5$ and $x=3$.

Example 4: Solve for $m$, $c=\dfrac{a}{b(m+n)}$.

Solution:

First, cross-multiply:

$c\cdot b(m+n)=a$

Now isolate $m+n$ by dividing by $cb$:

$m+n=\dfrac{a}{cb}$

Subtract $n$:

$\boxed{m=\dfrac{a}{cb}-n}$

Example 1: Solve for $x$, $\dfrac{3}{x}=\dfrac{8}{7}$

Solution: Cross-multiplying,

$3\cdot7=8\cdot x$

Dividing by $8$,

$\dfrac{21}{8}=x$

That is, $\boxed{x=\dfrac{21}{8}}$

Example 2: Solve for $y$, $\dfrac{4}{5}=\dfrac{7}{y+2}$

Solution: Cross-multiplying,

$4\cdot (y+2)=5\cdot 7$

Dividing by $4$,

$y+2=\dfrac{35}{4}$

or

$y=\dfrac{35}{4}-2$

or

$y=\dfrac{27}{4}$

Example 3: Solve, $\dfrac{3}{x}=\dfrac{-4}{x-7}$

Solution: Cross-multiplying,

$3(x-7)=-4x$

$3x-21=-4x$

or

$x=3$.