Elite Mathematics / Pure 1 / Topics / Simultaneous Equations IAL · WMA11 · Topic Simultaneous Equations: past paper questions & worked solutions
Every Simultaneous Equations question from the WMA11 papers in our bank, 2023–2026, sorted from easiest to hardest. Try each one first, then open the worked solution.
3 questions
18 marks in total
16% of papers (3 of 19) include it
2/1/0 easy / medium / hard
Questions, easiest first Level is estimated from marks: 1–6 easy, 7–8 medium, 9+ hard.
1. WMA11/01 January 2025 · Q3 Easy 5 marksShow worked solution Use t equals 5 The population is $58$ thousand, so $58^2=a+b(5^3)$, giving $3364=a+125b$.
Use t equals 10 The population is $65$ thousand, so $65^2=a+b(10^3)$, giving $4225=a+1000b$.
Solve for b Subtracting gives $861=875b$, so $b=\dfrac{861}{875}=\dfrac{123}{125}$.
Solve for a $a=3364-125\left(\dfrac{123}{125}\right)=3241$.
Use population 85 thousand $85^2=3241+\dfrac{123}{125}T^3$.
Find T $T^3=\dfrac{125(7225-3241)}{123}$, so $T\approx15.938$ and $T=15.9$ to one decimal place.
Final answer $(a)\ a=3241,\ b=\dfrac{123}{125}.\quad (b)\ T=15.9$ years.
2. WMA11/01 January 2026 · Q2 Easy 6 marksShow worked solution Use the linear equation $y-x=4$, so $y=x+4$.
Substitute into the second equation $2x^2-xy=8$ becomes $2x^2-x(x+4)=8$.
Show the quadratic $2x^2-x^2-4x=8$, so $x^2-4x-8=0$.
Complete the square $x^2-4x-8=0$ gives $(x-2)^2=12$.
Find x $x=2\pm\sqrt{12}=2\pm2\sqrt3$.
Find y Since $y=x+4$, $y=6\pm2\sqrt3$ with matching signs.
State both solutions $(x,y)=(2+2\sqrt3,6+2\sqrt3)$ or $(2-2\sqrt3,6-2\sqrt3)$.
Final answer $(x,y)=(2+2\sqrt3,\ 6+2\sqrt3)$ or $(2-2\sqrt3,\ 6-2\sqrt3)$.
3. WMA11/01 October 2023 · Q8 Medium 7 marksShow worked solution Write C1 in terms of y $xy=\dfrac{15}{2}-5x$, so $y=\dfrac{15}{2x}-5$.
Equate the two curves $\dfrac{15}{2x}-5=x^3-\dfrac72x-5$.
Multiply by 2x $15=2x^4-7x^2$, hence $2x^4-7x^2-15=0$.
Solve for x Let $u=x^2$. Then $2u^2-7u-15=0=(2u+3)(u-5)$, so $x^2=5$ and $x=\pm\sqrt5$.
Find the y-coordinates Using $y=\dfrac{15}{2x}-5$, the points are $\left(\sqrt5,\dfrac{3\sqrt5}{2}-5\right)$ and $\left(-\sqrt5,-\dfrac{3\sqrt5}{2}-5\right)$.
Use the distance formula $\Delta x=2\sqrt5$ and $\Delta y=3\sqrt5$, so $PQ=\sqrt{(2\sqrt5)^2+(3\sqrt5)^2}=\sqrt{65}$.
Final answer $(a)\ 2x^4-7x^2-15=0.\quad (b)\ PQ=\sqrt{65}$.
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