ONE EIGHTH CONTEST 2025
Anlo SHS - 31 points
Mpraeso SHS - 35 points
Mfantsiman Girls' SHS - 27 points
ONE EIGHTH CONTEST 2025
Pope John SHS & Minor Seminary - 58 points
Ghana SHS, Tamale - 25 points
Awudome SHS - 23 points
ONE EIGHTH CONTEST 2025
Swedru Secondary School - 36 points
Kumasi High School - 31 points
Anlo SHS - 30 points
ROUND 1
PREAMBLE
Find the solution of the given rational inequality
FIRST QUESTION
$x$ on the expression $x-2$ is less than zero: $\frac{x}{x-2}<0$
SOLUTION
Identify the critical points.
Equate both the numerator and denominator to zero.
$x=0$;
$x-2=0$ $x=2$.
The critical points are $0$ and $2$
Test with a number between 0 and 2.
We pick 1 to test if we will get a number less than zero when we substitute into $\frac{x}{x-2}<0$
$\frac{1}{1-2}<0$
$\frac{1}{-1}<0$
$-1<0$
It satisfies the inequality therefore set $x$ such that zero less than $x$ less than $2$ $(x:0<x<2)$
SECOND QUESTION
The expression $x+1$ on $x$ is greater than zero: $\frac{x+1}{x}>0$
SOLUTION
Identify the critical points.
Equate both the numerator and denominator to zero
$x+1=0$
$x=-1$;
$x=0$.
The critical points are $-1$ and $0$
Test with a number outside -1 and 0.
We want to pick a whole number to test.
We pick 1 to test: $\frac{1+1}{1}>0$ $2>0$
When we tested with 1 it was able to satisfy the inequality, so the solution does not lie between -1 and 0.
Therefore set $x$ such that x is less than -1 or x is greater than zero:
$(x:x<-1$ or $x>0)$
THIRD QUESTION
The expression $x+2$ on the expression $x-3$ is less than zero:
$\frac{x+2}{x-3}<0$
SOLUTION
Identify the critical points.
Equate both the numerator and denominator to zero
$x+2=0$
$x=-2$;
$x-3=0$
$x=3$.
The critical points are $-2$ and $3$.
Test with a number between -2 and 3.
We pick 0: $\frac{0+2}{0-3}<0$ $\frac{2}{-3}<0$.
It satisfies the inequality therefore set $x$ such that -2 less than $x$ less than $3$: $(x:-2<x<3)$
PRACTICE QUESTIONS
1. Find the solution of the rational inequality $\dfrac{x}{x-2}\lt0$.
ANSWER: $0\lt x\lt 2$
SOLUTION
Find the critical points where the numerator or denominator is zero:
$x=0\Rightarrow x=0$
$x-2=0\Rightarrow x=2$
Test a value in each region (positive, negative, positive for the three regions in order):
$0\lt x\lt 2$
2. Find the solution of the rational inequality $\dfrac{x-3}{x+1}\gt0$.
ANSWER: $x\lt -1\text{ or }x\gt 3$
SOLUTION
Find the critical points where the numerator or denominator is zero:
$x-3=0\Rightarrow x=3$
$x+1=0\Rightarrow x=-1$
Test a value in each region (positive, negative, positive for the three regions in order):
$x\lt -1\text{ or }x\gt 3$
3. Find the solution of the rational inequality $\dfrac{2x+1}{x-4}\lt0$.
ANSWER: $-\dfrac{1}{2}\lt x\lt 4$
SOLUTION
Find the critical points where the numerator or denominator is zero:
$2x+1=0\Rightarrow x=-\dfrac{1}{2}$
$x-4=0\Rightarrow x=4$
Test a value in each region (positive, negative, positive for the three regions in order):
$-\dfrac{1}{2}\lt x\lt 4$
4. Find the solution of the rational inequality $\dfrac{x+2}{2x-1}\gt0$.
ANSWER: $x\lt -2\text{ or }x\gt \dfrac{1}{2}$
SOLUTION
Find the critical points where the numerator or denominator is zero:
$x+2=0\Rightarrow x=-2$
$2x-1=0\Rightarrow x=\dfrac{1}{2}$
Test a value in each region (positive, negative, positive for the three regions in order):
$x\lt -2\text{ or }x\gt \dfrac{1}{2}$
5. Find the solution of the rational inequality $\dfrac{3x-6}{x+3}\lt0$.
ANSWER: $-3\lt x\lt 2$
SOLUTION
Find the critical points where the numerator or denominator is zero:
$3x-6=0\Rightarrow x=2$
$x+3=0\Rightarrow x=-3$
Test a value in each region (positive, negative, positive for the three regions in order):
$-3\lt x\lt 2$
6. Find the solution of the rational inequality $\dfrac{x-5}{x+2}\leq0$.
ANSWER: $-2\lt x\leq 5$
SOLUTION
Find the critical points where the numerator or denominator is zero:
$x-5=0\Rightarrow x=5$
$x+2=0\Rightarrow x=-2$
Test a value in each region (positive, negative, positive for the three regions in order):
The numerator root is included; the denominator root is never included.
$-2\lt x\leq 5$