# If 2((3x+ 1)/(2x- 5))^2 + (3x + 1)/(2x- 5) - 6= 0, then what is the value of x?

Apr 10, 2017

$x = \frac{9}{7}$

#### Explanation:

Let $u = \frac{3 x + 1}{2 x - 5}$. Then the equation becomes:

$2 {u}^{2} + u - 6 = 0$

$2 {u}^{2} + 4 u - 3 u - 6 = 0$

$2 u \left(u + 2\right) - 3 \left(u + 2\right) = 0$

$\left(2 u - 3\right) \left(u + 2\right) = 0$

$u = \frac{3}{2} \mathmr{and} - 2$

We now return to our original variable $x$.

Case 1: $u = \frac{3}{2}$

$\frac{3}{2} = \frac{3 x + 1}{2 x - 5}$

$3 \left(2 x - 5\right) = 2 \left(3 x + 1\right)$

$6 x - 15 = 6 x + 2$

$0 x = 17$

$x = \emptyset$

Case 2: $u = - 2$

$- 2 = \frac{3 x + 1}{2 x - 5}$

$- 2 \left(2 x - 5\right) = 3 x + 1$

$- 4 x + 10 = 3 x + 1$

$9 = 7 x$

$x = \frac{9}{7}$

Hopefully this helps!

Apr 10, 2017

$x = \frac{9}{7}$

#### Explanation:

For starters, you know that $x$ cannot be equal to $\frac{5}{2}$ because that would make the two denominators equal to zero.

In other words, you need to have

$2 x - 5 \ne 0$

$x \ne \frac{5}{2}$

Now, let's say that

(3x+ 1)/(2x - 5) = y" "color(darkorange)("(*)")

The original equation becomes

$2 {y}^{2} + y - 6 = 0$

You can solve this quadratic equation by using the quadratic formula, which for a general-form quadratic equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{a {x}^{2} + b x + c = 0}}}$

looks like this

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 \cdot a \cdot c}}{2 \cdot a}}}}$

$\left\{\begin{matrix}a = 2 \\ b = 1 \\ c = - 6\end{matrix}\right.$

and so

${y}_{1 , 2} = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \cdot 2 \cdot \left(- 6\right)}}{2 \cdot 2}$

${y}_{1 , 2} = \frac{- 1 \pm \sqrt{1 + 48}}{4}$

${y}_{1 , 2} = \frac{- 1 \pm \sqrt{49}}{4} \implies \left\{\begin{matrix}{y}_{1} = \frac{- 1 - 7}{4} = - 2 \\ {y}_{2} = \frac{- 1 + 7}{4} = \frac{3}{2}\end{matrix}\right.$

Take both values of $y$ and substitute them back into $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ to get

• $\underline{\text{For} \textcolor{w h i t e}{.} y = - 2}$

$\frac{3 x + 1}{2 x - 5} = - 2$

This is equivalent to

$3 x + 1 = - 2 \cdot \left(2 x - 5\right)$

$3 x + 1 = - 4 x + 10$

which gets you

$7 x = 9 \implies x = \frac{9}{7}$

• $\underline{\text{For} \textcolor{w h i t e}{.} y = \frac{3}{2}}$

$\frac{3 x + 1}{2 x - 5} = \frac{3}{2}$

This is equivalent to

$3 x + 1 = \frac{3}{2} \cdot \left(2 x - 5\right)$

$\textcolor{red}{\cancel{\textcolor{b l a c k}{3 x}}} + 1 = \textcolor{red}{\cancel{\textcolor{b l a c k}{3 x}}} - \frac{15}{2}$

$1 \ne - \frac{15}{2}$

Therefore, you can say that the original equation has $1$ valid solution, $x = \frac{9}{7}$. Notice that this solution satisfies the initial condition.

Do a quick double-check to make sure that the calculations are correct

$2 \cdot {\left(\frac{3 \cdot \frac{9}{7} + 1}{2 \cdot \frac{9}{7} - 5}\right)}^{2} + \frac{3 \cdot \frac{9}{7} + 1}{2 \cdot \frac{9}{7} - 5} - 6 = 0$

$2 \cdot {\left(\frac{\frac{27 + 7}{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}}}}{\frac{18 - 35}{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}}}}\right)}^{2} + \frac{\frac{27 + 7}{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}}}}{\frac{18 - 35}{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}}}} - 6 = 0$

$2 \cdot {\left(\frac{34}{- 17}\right)}^{2} + \left(\frac{34}{- 17}\right) - 6 = 0$

$2 \cdot {\left(- 2\right)}^{2} + \left(- 2\right) - 6 = 0$

$2 \cdot 4 - 8 = 0 \text{ } \textcolor{\mathrm{da} r k g r e e n}{\sqrt{}}$