ONE EIGHTH CONTEST 2025
Amaniampong SHS - 55 points
Aburi Presby SHS - 35 points
Adiembra SHS - 28 points
ONE EIGHTH CONTEST 2025
Osei-Tutu SHS - 43 points
West Africa SHS - 36 points
Hohoe E.P - 21 points
ONE EIGHTH CONTEST 2025
St James Seminary SHS - 54 points
Mankessim SHTS - 38 points
Presby Suhum - 16 points
ROUND 1
Solve for x given:
FIRST QUESTION
The absolute value of the expression $2x-3$ is equal to $x$: $|2x-3|=x$
ANSWER: $x = 3$ or $x = 1$
SOLUTION 1
For equations like $|2x-b|=x$, the solutions are always:
$x = b$
$x = \dfrac{b}{3}$
Given $b = 3$:
$x = 3$
$x = \dfrac{3}{3}$
$x = 1$
SOLUTION 2
Instead of squaring or multiplying cases, open the absolute value signs directly as a plus-or-minus option:
$2x - 3 = \pm x$
Case 1 (Plus):
$2x - 3 = x$
$2x - x = 3$
$x = 3$
Case 2 (Minus):
$2x - 3 = -x$
$2x + x = 3$
$3x = 3$
$x = 1$
SECOND QUESTION
The absolute value of the expression $2x-6$ is equal to $x$: $|2x-6|=x$
ANSWER: $x = 6$ or $x = 2$
SOLUTION 1
Given $b = 6$:
$x = 6$
$x = \dfrac{6}{3}$
$x = 2$
SOLUTION 2
$2x - 6 = \pm x$
Case 1 (Plus):
$2x - 6 = x$
$2x - x = 6$
$x = 6$
Case 2 (Minus):
$2x - 6 = -x$
$2x + x = 6$
$3x = 6$
$x = 2$
THIRD QUESTION
The absolute value of the expression $2x-12$ is equal to $x$: $|2x-12|=x$
ANSWER: $x = 12$ or $x = 4$
SOLUTION 1
Given $b = 12$:
$x = 12$
$x = \dfrac{12}{3}$
$x = 4$
SOLUTION 2
$2x - 12 = \pm x$
Case 1 (Plus):
$2x - 12 = x$
$2x - x = 12$
$x = 12$
Case 2 (Minus):
$2x - 12 = -x$
$2x + x = 12$
$3x = 12$
$x = 4$
PRACTICE QUESTIONS
1. Solve $|3x-2|=x$.
ANSWER: $x=1$ or $x=\dfrac{1}{2}$
SOLUTION
Since $|3x-2|=x$, we need $x\geq0$. Split into two cases:
Case 1: $3x-2=x$
$2x=2$
$x=1$
Case 2: $3x-2=-x$
$4x=2$
$x=\dfrac{1}{2}$
$x=1$ is not negative, so it is valid.
$x=\dfrac{1}{2}$ is not negative, so it is valid.
2. Solve $|2x-6|=x$.
ANSWER: $x=6$ or $x=2$
SOLUTION
Since $|2x-6|=x$, we need $x\geq0$. Split into two cases:
Case 1: $2x-6=x$
$1x=6$
$x=6$
Case 2: $2x-6=-x$
$3x=6$
$x=2$
$x=6$ is not negative, so it is valid.
$x=2$ is not negative, so it is valid.
3. Solve $|4x-9|=x$.
ANSWER: $x=3$ or $x=\dfrac{9}{5}$
SOLUTION
Since $|4x-9|=x$, we need $x\geq0$. Split into two cases:
Case 1: $4x-9=x$
$3x=9$
$x=3$
Case 2: $4x-9=-x$
$5x=9$
$x=\dfrac{9}{5}$
$x=3$ is not negative, so it is valid.
$x=\dfrac{9}{5}$ is not negative, so it is valid.
4. Solve $|5x-4|=x$.
ANSWER: $x=1$ or $x=\dfrac{2}{3}$
SOLUTION
Since $|5x-4|=x$, we need $x\geq0$. Split into two cases:
Case 1: $5x-4=x$
$4x=4$
$x=1$
Case 2: $5x-4=-x$
$6x=4$
$x=\dfrac{2}{3}$
$x=1$ is not negative, so it is valid.
$x=\dfrac{2}{3}$ is not negative, so it is valid.
5. Solve $|3x+4|=x$.
ANSWER: No solution
SOLUTION
Since $|3x+4|=x$, we need $x\geq0$. Split into two cases:
Case 1: $3x+4=x$
$2x=-4$
$x=-2$
Case 2: $3x+4=-x$
$4x=-4$
$x=-1$
$x=-2$ is negative, so it is rejected.
$x=-1$ is negative, so it is rejected.
6. Solve $|2x-5|=x$.
ANSWER: $x=5$ or $x=\dfrac{5}{3}$
SOLUTION
Since $|2x-5|=x$, we need $x\geq0$. Split into two cases:
Case 1: $2x-5=x$
$1x=5$
$x=5$
Case 2: $2x-5=-x$
$3x=5$
$x=\dfrac{5}{3}$
$x=5$ is not negative, so it is valid.
$x=\dfrac{5}{3}$ is not negative, so it is valid.