Higher November 2019 Paper 1 Q18

18 Solve \(\quad \dfrac{x + 15}{3} = 2(x + 10)\) [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(2x + 20\)	M1	correct expansion
\(x + 15 = 6x + 60\)	M1dep	multiplication by 3
\(15 - 60 = 6x - x\) or \(-45 = 5x\) or \(60 - 15 = x - 6x\) or \(45 = -5x\)	M1dep	collects terms
\(-9\)	A1	SC2 \(-3\) from \(2x + 10\) or 1 from \(6x + 10\)
Alternative method 2
\(2x + 20\)	M1	correct expansion
\(\dfrac{x}{3} + 5 = 2x + 20\) and \(5 - 20 = 2x - \dfrac{x}{3}\) or \(-15 = \dfrac{5x}{3}\) or \(20 - 5 = \dfrac{x}{3} - 2x\) or \(15 = -\dfrac{5x}{3}\)	M1dep	splits the fraction and collects terms
\(15 - 60 = 6x - x\) or \(-45 = 5x\) or \(60 - 15 = x - 6x\) or \(45 = -5x\)	M1dep	multiplication by 3
\(-9\)	A1	SC2 \(-3\) from \(2x + 10\) or 1 from \(6x + 10\)
Alternative method 3
\(6(x + 10)\) or \(6x + 60\)	M1	multiplication of rhs by 3
\(x + 15 = 6x + 60\)	M1dep	correct expansion
\(15 - 60 = 6x - x\) or \(-45 = 5x\) or \(60 - 15 = x - 6x\) or \(45 = -5x\)	M1dep	collects terms
\(-9\)	A1	SC2 \(-3\) from \(2x + 10\) or 1 from \(6x + 10\)