# How do solve 1/4<7/(7-x) algebraically?

Sep 11, 2017

$- 21 < x < 7$

#### Explanation:

We seek $x$ such that:

$\frac{1}{4} < \frac{7}{7 - x}$

The steps to solving any nonlinear inequality are:

• Rewrite the inequality so that there is a zero on the right side, in the form $f \left(x\right) < 0$ or $f \left(x\right) > 0$ (or $f \left(x\right) \le 0$ or $f \left(x\right) \ge 0$).
• Factorise the function, $f \left(x\right)$ if possible.
• Find the critical values where a sign change can occur by finding the roots of $f \left(x\right) = 0$, (or any factor of $f \left(x\right) = 0$).
• Determine the sign of $f \left(x\right)$ in the intervals formed by the critical values.

The solution will be those intervals in which the function has the correct signs satisfying the inequality.

We can rearrange the equation as follows:

$\frac{1}{4} < \frac{7}{7 - x} \implies \frac{1}{4} - \frac{7}{7 - x} < 0$

$\therefore \frac{1 \left(7 - x\right) - 7 \left(4\right)}{4 \left(7 - x\right)} < 0$

$\therefore \frac{7 - x - 28}{7 - x} < 0$

$\therefore \frac{- x - 21}{7 - x} < 0$

$\therefore \frac{x + 21}{x - 7} < 0$

Which we can write as:

$f \left(x\right) < 0$ where $f \left(x\right) = \frac{x + 21}{x - 7}$

So, the critical points where a sign change can occur, are:

$x + 21 = 0 \implies x = - 21$
$x - 7 = 0 \setminus \setminus \implies x = 7$

We must now examine the sign of $f \left(x\right)$ in each interval, partitioned by these critical points i.e:

 x lt -21; -21 lt x lt 7; x gt 7

The easiest way to do this is to look at the sign of each factor (or component) of $f \left(x\right)$ via a sign chart

 {: ( ul("factor"), ul(x lt -21), ul(-21 lt x lt 7), ul(x gt 7) ), ( x+21, -, +, + ), ( x-7, -, -, + ), ( { x + 21 } / { x-7 }, +,- ,+ ) :}

So, overall, we have established that the function, $f \left(x\right) < 0$ if $- 21 < x < 7$, so the solution of the inequality is:

$- 21 < x < 7$