5.12H Solving Simultaneous Equations Graphically: Higher Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 75 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Two straight lines are given by y = 2x - 3 and y = -x + 4. Work out the coordinates of their intersection point.

(Total for Question 1 is 2 marks)

A line with equation y = 0.5x + 1 and a parabola with equation y = x² - 2x + 2 are drawn on the same axes. Work out the x-coordinates of their points of intersection.

(Total for Question 2 is 3 marks)

Sketch on the same axes the graphs of y = -x + 6 and y = x² - 4x + 3 and use your sketch to estimate the x-coordinates of their intersections. Give your answers to 1 decimal place.

(Total for Question 3 is 3 marks)

The graph of y = kx + 2 meets the curve y = 3 - x² at exactly one point. Find the value(s) of k.

(Total for Question 4 is 4 marks)

A straight line passes through the points where y = x² - 5x + 4 crosses the x-axis. Find the equation of this straight line and hence find its intersections with the curve y = x² - 5x + 4 again, confirming consistency.

(Total for Question 5 is 4 marks)

Show that the graphs of y = 2x + 1 and y = -x² + 5x - 1 intersect at x = 1 and x = 2. Then find the y-coordinates of the intersection points and give the coordinates.

(Total for Question 6 is 5 marks)

The straight line L has equation y = mx + 3 and is tangent to the curve y = x² - 2x + 7 at x = a. Using the condition that the line and curve have equal gradients at the point of tangency and that they meet at x = a, express m in terms of a, and find the possible value(s) of a. Give the coordinates of the tangency point when a = 2.

(Total for Question 7 is 6 marks)

Consider the pair of equations y = 0.2x² + x - 1 and y = 0.8x + 2. Use algebra to form the equation whose roots are the x-coordinates of the intersections. Then find those x-coordinates to 2 decimal places and give the intersection points.

(Total for Question 8 is 3 marks)

Mark scheme · 5.12H Solving Simultaneous Equations Graphically: Higher Tier Practice

Question 1

M1 substitute or set 2x - 3 = -x + 4 and solve for x
A1 x = 7/3, y = 5/3 cao
Answer:(7/3, 5/3)
(Approximately (2.333, 1.667))

Question 2

M1 equate x² - 2x + 2 = 0.5x + 1 to form quadratic x² - 2.5x + 1 = 0
M1 use quadratic formula or complete the square to find x
A1 x = (2.5 ± √2.25)/2 = (2.5 ± 1.5)/2 so x = 2 or x = 0.5 cao
Answer:x = 2 and x = 0.5
(Points of intersection at x = 2 and x = 1/2)

Question 3

M1 sketch showing both graphs with at least one intersection roughly located and correct axes scale
M1 identify two intersection x-values from sketch within reasonable accuracy
A1 answers x approx -0.8 and x approx 3.8 to 1 dp (or equivalent to 1 dp) cao
Answer: x = -0.8 and x = 3.8 (to 1 decimal place)

Question 4

M1 set kx + 2 = 3 - x² to form x² + kx -1 = 0 and state condition for one root is discriminant = 0
M1 compute discriminant: k² - 4(1)(-1) = k² + 4
M1 set k² + 4 = 0 and conclude no real k makes discriminant zero, or recognise discriminant always positive
A1 no real value of k gives exactly one real intersection; always two real intersections for real k cao
Answer: No real value of k. The discriminant k² + 4 is always positive for real k, so the quadratic has two distinct real roots for every real k.

Question 5

M1 find x-intercepts of curve: solve x² - 5x + 4 = 0 giving x = 1 and x = 4
M1 find equation of line through (1,0) and (4,0): recognise it is the x-axis y = 0
M1 substitute y = 0 into curve to verify intersections at x = 1 and x = 4
A1 equation y = 0 and intersections x = 1, 4 cao
Answer: Equation: y = 0. Intersections with the curve occur at x = 1 and x = 4 (points (1,0) and (4,0)).

Question 6

M1 equate 2x + 1 = -x² + 5x - 1 to get x² - 3x + 2 = 0 and show factor or algebraic manipulation leading to (x-1)(x-2)=0 or show substitution gives zero
M1 obtain x = 1 and x = 2 (show roots) oe
M1 substitute x = 1 into y = 2x + 1 to get y = 3, and x = 2 gives y = 5
A1 coordinates (1, 3) and (2, 5) cao
B1 explicit statement 'hence' linking substitution to intersection coordinates
Answer: Intersection points (1, 3) and (2, 5).

Question 7

M1 state gradient of curve dy/dx = 2x - 2 and set m = 2a - 2 for tangent at x = a
M1 state point of contact satisfies a² - 2a + 7 = m a + 3
M1 substitute m = 2a - 2 into equation to get a² - 2a + 7 = (2a - 2)a + 3 = 2a² - 2a + 3
M1 rearrange correctly to -a² + 4 = 0, i.e. a² = 4, giving a = 2 or a = -2
A1 state a = 2 gives m = 2(2) - 2 = 2, and a = -2 gives m = 2(-2) - 2 = -6 cao
A1 for given a = 2, compute coordinates: y = 2² - 2(2) + 7 = 7, so tangency point (2, 7), with working shown cao
Answer: m = 2a - 2; tangency occurs when a = 2 (m = 2) or a = -2 (m = -6). For a = 2 the tangency point is (2, 7).

Question 8

M1 equate 0.2x² + x -1 = 0.8x + 2 to form 0.2x² + 0.2x -3 = 0 or multiply to x² + x -15 = 0
M1 use quadratic formula to find x = [-1 ± √1 + 60]/2 = [-1 ± √61]/2
A1 x approx ( -1 + 7.8102)/2 = 3.4051 -> 3.41 and x approx ( -1 - 7.8102)/2 = -4.4051 -> -4.41; intersection points (3.41, y) and (-4.41, y) cao
Answer: x = ( -1 ± √61 )/2 approx 3.41 and -4.41. Intersection points approx (3.41, 0.8*3.41 + 2 = 4.73) and (-4.41, 0.8*(-4.41) + 2 = -1.53) to 2 dp.

Solving Simultaneous Equations Graphically: Higher Tier Practice - Worksheets, Questions and Revision