Solving Simultaneous equations by the Graphical Method

Simultaneous linear equations Solve . y = x + 2 y = 2x + 5 y = x + 2 y = x + 2 y = 2x + 5 y = 2x + 5 From the graph, the solution of the above simultaneous equations is x = 1, y = 3. (1, 3) Solution

Similarly, we can solve simultaneous equations with one linear and one quadratic using the graphical method.

The solutions obtained are approximations only. Simultaneous equations (one linear and one quadratic) y = x2 Solve . y = x2 y = x + 2 y = x2 y = x + 2 Solution y = x + 2 (2, 4) Referring to the figure, the two graphs intersect at (1, 1) and (2, 4). Solution (1, 1) The solutions obtained are approximations only.

Solve the following simultaneous equations graphically. y = x2  x + 3 y = 2x + 3 Step 1 Draw the graphs of both equations on the same rectangular coordinate plane. y = x2  x + 3 x 2 1 1 2 3 y 9 5 y = 2x + 3 x 1 2 y 3 5 7

Read the coordinates of the intersection(s) of their graphs. x y 2 1 1 2 3 10 8 6 4 2 Step 2 Read the coordinates of the intersection(s) of their graphs. (3, 9) ∵ The two graphs intersect at (0, 3) and (3, 9). y = x2  x + 3 (0, 3) y = 2x + 3 ∴ (x, y) = (0, 3) or (3, 9)

If the two graphs do not intersect, what are the solutions? y x If the two graphs do not intersect, the simultaneous equations have no real solutions.

Follow-up question Solve the following simultaneous equations graphically. y = x2  3x + 1 y = x + 5 y = x2  3x + 1 x 4 3 2 1 1 y 3 y = x + 5 x 4 3 2 y 1 2 3

∵ The two graphs intersect at only one point (2, 3). 1 1 2 3 (2, 3) ∴ (x, y) = (2, 3) y = x + 5 x 4 3 2 1 1 y = x2  3x + 1

4 2 y Do you know how many solutions do this set of simultaneous equations have? If you extend the range of the graphs, you will find another intersection! x 2 1 1 2 As you see, the two graphs intersect at only one point, the simultaneous equations have only one solution.

Yes, we can find the number of solutions by algebraic method. Without drawing the graphs of the simultaneous equations, can we find the number of their solutions? Yes, we can find the number of solutions by algebraic method.

Consider the following simultaneous equations. y = ax2 + bx + c ……(1) y = mx + n ……(2) By substituting (2) into (1), we have mx + n = ax2 + bx + c ax2 + (b  m)x + (c  n) = 0 ……(3) and also the number of intersections of their graphs by considering the discriminant () of the equation (3). We can determine the number of real solutions of the simultaneous equations,

2 distinct real solutions Discriminant () of ax2 + (b  m)x + (c  n) = 0 No. of solutions of y = ax2 + bx + c y = mx + n No. of intersections of the graphs of y = ax2 + bx + c and y = mx + n  > 0  = 0  < 0 2 2 distinct real solutions 1 real solution 1 no real solutions no no no intersections e.g. y x y x 2 intersections e.g. 2 1 intersection e.g. y x 1

Follow-up question Find the number of intersections of the line L: y = 2x – 4 and the quadratic curve C: y = 3x2 – 6x – 2. y = 3x2 – 6x – 2 ……(1) y = 2x – 4 ……(2) By substituting (2) into (1), we have 2 6 3 4 - = x 2 8 3 = + - x ……(3) of (3) = (–8)2 – 4(3)(2) = 40 > 0

Follow-up question Find the number of intersections of the line L: y = 2x – 4 and the quadratic curve C: y = 3x2 – 6x – 2. ∵  of (3) > 0 ∴ (3) has 2 distinct real roots. ∴ The simultaneous equations have 2 distinct real solutions. ∴ There are 2 intersections between the two graphs.