PRELIMINARY CONTEST 2024
Contest 4
Tsito SHTS – 55 pts (Winner)
Methodist Girls’ SHS, Mamfe – 48 pts
Notre Dame Seminary SHS – 45 pts
Contest 5
Tema Methodist SHS – 54 pts (Winner)
Shama SHS – 37 pts
Bolgatanga SHS – 28 pts
Contest 6
Wallahs Academy – 32 pts (Winner)
O’Reilly SHS – 29 pts
Adanwomase SHS – 25 pts
ROUND 1
FIRST SET OF QUESTIONS
Find an equation in $x$ and $y$ in the form $y = f(x)$ from the given equation
QUESTION
1. $x=2t+3$ and $y=t^{2}+5$
SOLUTION
$x=2t+3$
$t=\frac{x-3}{2}$
$y=t^{2}+5$
$y=(\frac{x-3}{2})^{2}+5$
2. $x=3t-5$ and $y=t^{2}-6$
SOLUTION
$x=3t-5$
$t=\frac{x+5}{3}$
$y=t^{2}-6$
$y=(\frac{x+5}{3})^{2}-6$
3. $x=4t-1 and y=t^{2}+2$
SOLUTION
$x=4t-1$
$t=\frac{x+1}{4}$
$y=t^{2}+2$
$y=(\frac{x+1}{4})^{2}+2$
PRACTICE QUESTIONS
1. Find an equation in $x$ and $y$ in the form $y=f(x)$ from $x=2t+3$ and $y=t^2+5$.
ANSWER: $y=\dfrac{(x-3)^2}{4}+5$
SOLUTION
Isolate $t$ from the $x$ equation:
$x-3=2t$
$t=\dfrac{x-3}{2}$
Substitute into the $y$ equation:
$y=\left(\dfrac{x-3}{2}\right)^2+5$
$y=\dfrac{(x-3)^2}{4}+5$
2. Find an equation in $x$ and $y$ in the form $y=f(x)$ from $x=3t-5$ and $y=t^2-6$.
ANSWER: $y=\dfrac{(x+5)^2}{9}-6$
SOLUTION
Isolate $t$ from the $x$ equation:
$x+5=3t$
$t=\dfrac{x+5}{3}$
Substitute into the $y$ equation:
$y=\left(\dfrac{x+5}{3}\right)^2-6$
$y=\dfrac{(x+5)^2}{9}-6$
3. Find an equation in $x$ and $y$ in the form $y=f(x)$ from $x=4t-1$ and $y=t^2+2$.
ANSWER: $y=\dfrac{(x+1)^2}{16}+2$
SOLUTION
Isolate $t$ from the $x$ equation:
$x+1=4t$
$t=\dfrac{x+1}{4}$
Substitute into the $y$ equation:
$y=\left(\dfrac{x+1}{4}\right)^2+2$
$y=\dfrac{(x+1)^2}{16}+2$
4. Find an equation in $x$ and $y$ in the form $y=f(x)$ from $x=5t+2$ and $y=t^2-3$.
ANSWER: $y=\dfrac{(x-2)^2}{25}-3$
SOLUTION
Isolate $t$ from the $x$ equation:
$x-2=5t$
$t=\dfrac{x-2}{5}$
Substitute into the $y$ equation:
$y=\left(\dfrac{x-2}{5}\right)^2-3$
$y=\dfrac{(x-2)^2}{25}-3$
5. Find an equation in $x$ and $y$ in the form $y=f(x)$ from $x=2t-7$ and $y=t^2+4$.
ANSWER: $y=\dfrac{(x+7)^2}{4}+4$
SOLUTION
Isolate $t$ from the $x$ equation:
$x+7=2t$
$t=\dfrac{x+7}{2}$
Substitute into the $y$ equation:
$y=\left(\dfrac{x+7}{2}\right)^2+4$
$y=\dfrac{(x+7)^2}{4}+4$
6. Find an equation in $x$ and $y$ in the form $y=f(x)$ from $x=3t+1$ and $y=t^2+8$.
ANSWER: $y=\dfrac{(x-1)^2}{9}+8$
SOLUTION
Isolate $t$ from the $x$ equation:
$x-1=3t$
$t=\dfrac{x-1}{3}$
Substitute into the $y$ equation:
$y=\left(\dfrac{x-1}{3}\right)^2+8$
$y=\dfrac{(x-1)^2}{9}+8$