4MA0 Higher January 2012 Paper 4 Q13

EdexcelHigherOld spec8 marksLinear Graphs & Gradients Simultaneous Equations

(a) Find the gradient of the line with equation \(3x + 4y = 10\) (3)

(b) Find the coordinates of the point of intersection of the line with equation \(3x + 4y = 10\) and the line with equation \(5x - 6y = 23\)
Show your working clearly. (5)

Mark scheme (a)

Scheme	Marks
\(4y = 10 - 3x\) or \(-4y = 3x - 10\)	M1
\(y = \dfrac{5}{2} - \dfrac{3}{4}x\) oe or \(y = \dfrac{10}{4} - \dfrac{3}{4}x\) oe or \(y = \dfrac{10 - 3x}{4}\) oe	M1
\(-\dfrac{3}{4}\)	A1
	(3)

Notes

M1: May be implied by second M1 or by \(y = -\frac{3}{4}x + c\) even if value of \(c\) is incorrect.
or finds coordinates of 2 points on the line eg (0, 2.5), \(x = 2\), \(y = 1\), table, diagram.

M1: or for clear attempt to evaluate \(\dfrac{\text{vert diff}}{\text{horiz diff}}\) for their pts

A1: Award 3 marks for correct answer if either first M1scored or no working shown.
SC If M0, award B1 for \(-\frac{3}{4}x\)

Apart from Questions 3, 13(b) and 17(f) (where the mark scheme states otherwise), the correct answer, unless clearly obtained by an incorrect method, should be taken to imply a correct method.

Mark scheme (b)

Scheme	Marks
eg \(9x + 12y = 30\), \(10x - 12y = 46\) or eg \(15x + 20y = 50\), \(15x - 18y = 69\)	M1
\(x = 4\) or \(y = -\frac{1}{2}\)	A1
	M1
\(x = 4\), \(y = -\frac{1}{2}\)	A1
\((4, -\frac{1}{2})\)	B1
	(5)
	(Total 8 marks)

Notes

M1: for coefficients of \(x\) or \(y\) the same or for correct rearrangement of one equation followed by correct substitution in the other

eg \(5x - 6\left(\dfrac{10 - 3x}{4}\right) = 23\)

A1: cao dep on M1

M1: (dep on 1st M1) for substituting for other variable

A1: Award 4 marks for correct values if at least first M1 scored

B1: Award 5 marks for correct answer if at least first M1 scored
ft from their values of \(x\) and \(y\)