Grade 10 Maths · Algebra

Solve a linear equation

Solve an equation by doing the same inverse operation on both sides until the unknown stands alone (for example 3x + 4 = 19 gives x = 5).

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Linear Equations Grade 9: IntroductionKevinmathscience

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Question 9.2 · November 2024 · Paper 2 · Provincial · 3 marks

In the diagram below, R is the midpoint of QT and S is the midpoint of PT. The length of SR is

2 x + 4

\hat{Q} = 3 9^{\circ}

and

\hat{P} = 5 5^{\circ}

9.2 If the length of PQ is 18, determine the value of x.(3)

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More past paper questions: solve a linear equation

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3. Question 1.1.3 · May/June 2025 · Paper 1 · National (DBE) · 3 marks
1.1 Solve for $x$ :
1.1.3 $2^{x + 4} + 2^{x} = 8704$ (3)
Show the memo (3 ticks)Memo
1. ✓
  Factorise the common exponential term $2^{x} (16 + 1) = 8704$
  1 mark · Take out a common power
2. ✓
  Simplify to a single exponential equation $2^{x} (17) = 8704$
  $2^{x} = 512$
  1 mark · Solve a linear equation
3. ✓
  Solve for x $2^{x} = 2^{9}$
  $x = 9$
  1 mark · Solve an exponential equation
4. Question 2.1.1 · March 2021 · Provincial · 2 marks
2.1 Solve for x in each of the following equations:
2.1.1 $1 - 5 x = 6$ (2)
Show the memo (2 ticks)Memo
1. ✓
  Simplification $- 5 x = 5$
  1 mark · Solve a linear equation
2. ✓
  Solve $x = - 1$
  1 mark · Solve a linear equation
5. Question 1.1.5 · May/June 2024 · Paper 1 · National (DBE) · 5 marks
1.1 Solve for x:
1.1.5 $\frac{x}{20 - x} = 1$ (5)
Show the memo (5 ticks)Memo
1. ✓
  Isolate the surd $x = 20 - x$
  1 mark · Solve a linear equation
2. ✓
  Square both sides $x^{2} = 20 - x$
  1 mark · Solve a surd equation
3. ✓
  Standard form $x^{2} + x - 20 = 0$
  1 mark · Write a quadratic in standard form
4. ✓
  Solve quadratic factors $x = 4$ or $x = - 5$
  1 mark · Solve a quadratic by factorising
5. ✓
  Select valid solution after check $x = 4$
  1 mark · Check solutions of a surd equation
6. Question 9.5 · May/June 2025 · Paper 1 · National (DBE) · 6 marks
Given: $f (x) = x^{3} - 6 x^{2} + 9 x - 4 = (x - 4) (x - k)^{2}$
9.5 Calculate the maximum vertical distance between $f$ and $h$ for $1 < x < 3$ , if $h (x) = - 2 f^{'} (x)$ .(6)
Show the memo (6 ticks)Memo
1. ✓
  Establish the expression for the distance $h (x) = - 2 (3 x^{2} - 12 x + 9) = - 6 x^{2} + 24 x - 18$
  $Distance = h (x) - f (x)$
  $= (- 6 x^{2} + 24 x - 18) - (x^{3} - 6 x^{2} + 9 x - 4)$
  1 mark · Find a horizontal or vertical length
2. ✓
  Simplify the distance expression $= - 6 x^{2} + 24 x - 18 - x^{3} + 6 x^{2} - 9 x + 4$
  $= - x^{3} + 15 x - 14$
  1 mark · Collect like terms
3. ✓
  Differentiate the distance expression and equate to 0To find max distance, let $D^{'} (x) = 0$ .
  $D^{'} (x) = - 3 x^{2} + 15 = 0$
  1 mark · Find stationary points
4. ✓
  Solve the derivative equation for x squared $3 x^{2} = 15$
  $x^{2} = 5$
  1 mark · Solve a linear equation
5. ✓
  Find the correct x-value within the given domain $x = \pm 5$
  But the domain is restricted to $1 < x < 3$ .
  $∴ x = 5$ (since $- 5$ is not in the interval)
  1 mark · Solve a quadratic by factorising
6. ✓
  Calculate the maximum vertical distanceSubstitute $x = 5$ into the distance equation:
  $Max distance = - (5)^{3} + 15 (5) - 14$
  $= - 55 + 155 - 14$
  $= 105 - 14 \approx 8, 36$
  1 mark · Find stationary points

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