Solving a rational equation with multiple and nested fractions

Question

This is the equation to solve: $\dfrac{\dfrac{x+\dfrac{1}{2}} {\dfrac{1}{2}+\dfrac{x}{3}}}{\dfrac{1}{4}+\dfrac{x}{5}}=3$

What I did:

$x+\dfrac{1}{2}=\dfrac{2x+1}{2}$

$\dfrac{x}{3}+\dfrac{1}{2}=\dfrac{2x+3}{6}$

$\dfrac{x}{5}+\dfrac{1}{4}=\dfrac{4x+5}{20}$

$\dfrac{2x+1}{2}\div \dfrac{2x+3}{6} = \dfrac{2x+1}{2}\times \dfrac{6}{2x+3}= \dfrac{6x+3}{2x+3}$

$\dfrac{6x+3}{2x+3}\div\dfrac{4x+5}{20}=\dfrac{6x+3}{2x+3}\times\dfrac{20}{4x+5}=3$

$\implies\dfrac{120x+60}{(2x+3)(4x+5)}=3$

I know how to solve the equation. But right now I tried several times and I got wrong answers.

So I appreciate your help. Thank you.

Answer 1

You have, essentially,

$$ \frac{120x+60}{(2x+3)(4x+5)} = 3 $$

Multiply both sides by $(2x+3)(4x+5)$ to get

$$ 120x+60 = 3(2x+3)(4x+5) = 24x^2+66x+45 $$

Subtract $120x+60$ from both sides to get

$$ 24x^2-54x-15 = 0 $$

Divide both sides by $3$ to get

$$ 8x^2-18x-5 = 0 $$

which factors as

$$ (4x+1)(2x-5) = 0 $$

to yield $x = -1/4$ or $5/2$. (Or, you can use the quadratic formula.)

Answer 2

$$\frac { \frac { 2x+1 }{ \frac { 2 }{ \frac { 3+2x }{ 6 } } } }{ \frac { 5+4x }{ 20 } } =3\quad \Rightarrow \frac { \frac { 2x+1 }{ 2 } \cdot \frac { 6 }{ 2x+3 } }{ \frac { 5+4x }{ 20 } } =3\quad \Rightarrow \frac { \frac { 2x+1 }{ 2x+3 } }{ \frac { 5+4x }{ 20 } } =1\Rightarrow \frac { 2x+1 }{ 2x+3 } =\frac { 5+4x }{ 20 } \Rightarrow \\ \Rightarrow 8{ x }^{ 2 }-18x-5=0\Rightarrow x=\frac { 9\pm 11 }{ 8 } $$

Answer 3

Notice, $$\frac{\frac{x+\frac{1}{2}}{\frac{1}{2}+\frac{x}{3}}}{\frac{1}{4}+\frac{x}{5}}=3$$ $$\frac{\frac{3(2x+1)}{(2x+3)}}{\frac{4x+5}{20}}=3$$ $$\frac{60(2x+1)}{(2x+3)(4x+5)}=3$$ $$8x^2-18x-5=0$$ $$x=\frac{18\pm\sqrt{(-18)^2-4(8)(-5)}}{2(8)}$$ $$x=\frac{18\pm22}{16}$$ $$x=\frac{5}{2}$$ or $$ x=-\frac{1}{4}$$