Elite Mathematics / Linear / Topics / Forming and Solving Equations IGCSE · 4MA1 · Topic Forming and Solving Equations: past paper questions & worked solutions
Every Forming and Solving Equations question from the 4MA1 papers in our bank, 2020–2025, sorted from easiest to hardest. Try each one first, then open the worked solution.
6 questions
25 marks in total
11% of papers (6 of 54) include it
0/4/2 easy / medium / hard
Questions, easiest first Level is estimated from marks: 1–2 easy, 3–4 medium, 5+ hard.
1. Nov 2025 P2H · Q20 Medium 3 marksShow worked solution Calculate value For
\[
ax^2+4x+c=0
\]
the quadratic formula gives \[
x=\frac{-4\pm \sqrt{4^2-4ac}}{2a}
\]
The given solutions are \[
x=\frac{-4\pm 2\sqrt{39}}{10}
\]
Compare denominators Compare denominators:
\[
2a=10
\]
\[
a=5
\]
Compare the square root parts Compare the square root parts:
\[
\sqrt{16-4ac}=2\sqrt{39}
\]
Square both sides Square both sides:
\[
16-4ac=156
\]
Use Use \(a=5\):
\[
16-20c=156
\]
\[
-20c=140
\]
\[
c=-7
\]
Final answer \(a=5,\ c=-7\)
2. May 2025 4WM1HR · Q6 Medium 3 marksShow worked solution Use trigonometry The base angles are equal, so the opposite sides are equal.
\[
4x+6=30-x
\]
\[
5x=24
\]
\[
x=4.8
\]
3. Nov 2021 P2H · Q2 Medium 4 marksShow worked solution Area of rectangle A Area of rectangle A:
\[
4(5-x)
\]
Area of rectangle B Area of rectangle B:
\[
5(2x-1)
\]
Rectangle B has twice the Rectangle B has twice the area of rectangle A:
\[
5(2x-1)=2\cdot 4(5-x)
\]
\[
10x-5=40-8x
\]
\[
18x=45
\]
\[
x=2.5
\]
4. Jan 2020 P2HR · Q3 Medium 4 marksShow worked solution Write each tin in terms of x Tin A has \(x\) buttons, tin B has \(4x\), tin C has \(x-7\).
Form and solve the equation \[
x+4x+(x-7)=137 \;\Rightarrow\; 6x-7=137 \;\Rightarrow\; 6x=144 \;\Rightarrow\; x=24
\]
Buttons in tin C \[
24-7=17
\]
Final answer \(17\) buttons
5. Nov 2021 P1H · Q5 Hard 5 marksShow worked solution The top row has 5 The top row has 5 tile widths and the bottom row has 4 tile lengths, so
\[
5W=4L
\]
\[
L=\frac54W
\]
Use trigonometry The rectangle has width \(5W\) and height \(L+W\):
\[
5W(L+W)=1620
\]
Substitute Substitute \(L=\frac54W\):
\[
5W\left(\frac54W+W\right)=1620
\]
\[
5W\left(\frac94W\right)=1620
\]
\[
\frac{45}{4}W^2=1620
\]
\[
W^2=144
\]
\[
W=12,\qquad L=\frac54(12)=15
\]
Final answer \(L=15\text{ cm}\), \(W=12\text{ cm}\).
6. Nov 2023 P1H · Q2 Hard 6 marksShow worked solution (a) \[
12pq-18p=6p(2q-3)
\]
Solve equation (b) The total number of bars is 56, so \[ y+(3y+7)+(2y-5)=56 \]
\[
6y+2=56
\]
\[
y=9
\]
The number of zinc bars The number of zinc bars is \[ 2y-5=2(9)-5=13 \]
Final answer (a) \(6p(2q-3)\), (b) \(13\) zinc bars
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