Grade 10 Maths · Algebra

Solve simultaneous equations by elimination

Solve two equations together by adding or subtracting them so one unknown cancels (for example x + y = 7 plus x − y = 1 gives 2x = 8).

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Simultaneous Equations grade 10 | EliminationKevinmathscience

Simultaneous Equations grade 10 | Introduction to Elimination and SubstitutionKevinmathscience

Exam Simultaneous Equations Grade 10Kevinmathscience

Method of Elimination Steps to Solve Simultaneous EquationsCowan Academy

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Question 8.3 · May/June 2025 · Paper 1 · National (DBE) · 6 marks

8.3 Given that

y = 4 x - 14

is a common tangent to

f (x) = 2 x^{2} - 4 x - 6

and

g (x) = a x^{2} + b x - 18

, calculate the values of

a

and

b

.(6)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: solve simultaneous equations by elimination

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3. Question 3.2.2(b) · May/June 2024 · Paper 1 · National (DBE) · 4 marks
3.2 Given an arithmetic sequence with $T_{1} = 8$ and $T_{2} = 11$ .
A new arithmetic sequence $P$ is formed using the term position and the term value of the given arithmetic sequence. For the new sequence, $P_{8} = 1$ , $P_{11} = 2$ and so forth.
3.2.2(b) Calculate the value of the first term of the new arithmetic sequence.(4)
Show the memo (4 ticks)Memo
1. ✓
  Set up first equation using given sequence data $a + 7 d = 1$
  1 mark · Find a linear general term
2. ✓
  Set up second equation using given sequence data $a + 10 d = 2$
  1 mark · Find a linear general term
3. ✓
  Solve for common difference (d) $3 d = 1 ⟹ d = \frac{1}{3}$
  1 mark · Solve simultaneous equations by elimination
4. ✓
  Solve for first term (a) $a + 7 (\frac{1}{3}) = 1 ⟹ a = - \frac{4}{3}$
  1 mark · Solve simultaneous equations by substitution
4. Question 2.3.3 · March 2018 · Provincial · 4 marks
2.3 Given: $4^{x + y} = 64$ and $3^{x - y - 1} = 1$
2.3.3 Hence or otherwise, find the value of $3^{x} \cdot 5^{y}$ (4)
Show the memo (3 ticks)Memo
1. ✓
  Solve simultaneous equations $x + y = 3$ ...(1)
  $x - y = 1$ ...(2)
  Adding (1) and (2): $2 x = 4 ⟹ x = 2$
  2 marks · Solve simultaneous equations by elimination
2. ✓
  Find y $2 + y = 3 ⟹ y = 1$
  1 mark · Solve simultaneous equations by substitution
3. ✓
  Calculate final expression $3^{x} \cdot 5^{y} = 3^{2} \cdot 5^{1}$
  $= 9 \cdot 5$
  $= 45$
  1 mark · Substitute into a formula
5. Question 4.1 · June 2024 · Paper 1 · Provincial · 4 marks
Consider the functions $f (x) = a x^{2} + c$ and $g (x) = b x + q$ . $f$ and $g$ intersect at points $A (- 1; - 3)$ and $B (2; 0)$ .
4.1 Calculate the values of a and c.(4)
Show the memo (3 ticks)Memo
1. ✓
  Substitute point A and B into equationSub A(-1; -3): $- 3 = a (- 1)^{2} + c ⟹ - 3 = a + c$ ... (1)
  Sub B(2; 0): $0 = a (2)^{2} + c ⟹ 0 = 4 a + c$ ... (2)
  2 marks · Find the equation of a parabola
2. ✓
  Solve simultaneously for aEq 1 - Eq 2:
  $- 3 - 0 = a - 4 a$
  $- 3 = - 3 a$
  $a = 1$
  1 mark · Solve simultaneous equations by elimination
3. ✓
  Solve for cSub $a = 1$ into Eq 1:
  $- 3 = 1 + c$
  $c = - 4$
  1 mark · Solve simultaneous equations by substitution
6. Question 8.3 · September 2024 · Paper 1 · Provincial · 6 marks
8.3 A function $g (x) = a x^{2} + \frac{b}{x}$ has a minimum value at $x = 4$ . The function value at $x = 4$ is 96. Calculate the values of a and b.(6)
Show the memo (6 ticks)Memo
1. ✓
  Derivative is 0 at the minimumMinimum at $x = 4$ , so $g^{'} (4) = 0$
  1 mark · Find stationary points
2. ✓
  Derivative $g (x) = a x^{2} + b x^{- 1}$ $g^{'} (x) = 2 a x - b x^{- 2}$
  1 mark · Differentiate using the power rule
3. ✓
  Equation 1 $2 a (4) - b (4)^{- 2} = 0$ $8 a - \frac{b}{16} = 0$ $128 a - b = 0$ ...(1)
  1 mark · Find stationary points
4. ✓
  Equation 2 $g (4) = 96$ : $16 a + \frac{b}{4} = 96$ $64 a + b = 384$ ...(2)
  1 mark · Substitute into a formula
5. ✓
  Value of a(1) + (2): $192 a = 384$ $a = 2$
  1 mark · Solve simultaneous equations by elimination
6. ✓
  Value of bFrom (1): $b = 128 (2) = 256$
  1 mark · Solve simultaneous equations by substitution

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