## Simultaneous Equations Simultaneous equations are systems of several equations with several variables.

The systems dealt with below are of two equations with two variables.

If the graphs of the two equations are drawn the solutions are where the graphs intersect.

### Two Linear Equations

These are usually first studied in Year 11.

There are four methods available at this level for solving simultaneous linear equations. The setting out for these problems needs to be done clearly and carefully, especially for the algebraic methods. Each equation should be labelled with a letter or number, and the solution set should be checked back into both equations.

 1. Graphical methods. Sketch the graph of each equation. Where they intersect is the solution set. This method is not very accurate and algebraic methods are often better used. Examples 2. Comparison method. If the two equations have the same variable as the subject of their equations, the other sides of the equations can be equated. Examples 3. Substitution method. If one of the equations has a variable alone as the subject, it can be directly substituted into the other equation. Examples 4. Elimination method. If the coefficients of either x or y are the same, or can easily be made the same, then either the x or the y term can be eliminated by adding or subtracting the two equations Examples

Linear equation and a Parabola

The line y = x and the parabola y = x2 + 8x + 10 intersect in two places (see graph below)

 To solve these two equations algebraically use the substitution method. y = x                    ...............Ay = x2 + 8x + 10 ...............B x = x2 + 8x + 10       (subst y = x into B) x2 + 7x + 10 = 0   (re-arranging) (x + 5)(x + 2) = 0  (factorising) x = -5 or x = -2 When x = -5, y = -5 (substituting into A) When x = -2, y = -2 (substituting into A) The solutions are (-5, -5) and (-2, -2) Linear equation and a Circle

The line y = 2x + 1 and the circle x2 + y2 = 1 intersect in two places (see graph below)

 To solve these two equations algebraically use the substitution method. y = 2x + 1    ...............Ax2 + y2 = 1   ...............B x2 + (2x + 1)2  = 1    (subst y = x into B) x2 + 4x2 + 4x + 1 = 1 (expanding) 5x2 + 4x = 0  (re-arranging) x(5x + 4) = 0 x =0 or x = -0.8 When x = 0 y = 2 x 0 + 1 = 1          (substituting into A) When x = -0.8y = 2 x − 0.8 + 1 = -0.6 (substituting into A) The solutions are (0, 1) and (-0.8, -0.6) Linear equation and a Hyperbola

The line 2x + y = 7 and the hyperbola xy = 5 intersect in two places (see graph below) To solve these two equations algebraically use the substitution method. y = 7 − 2x                     ...............Axy = 5                          ...............B x(7 − 2x) = 5         (subst y from A into B) 7x − 2x2 = 5               (expanding) -2x2 + 7x − 5 = 0       (re-arranging) (-2x + 5)(x − 1) = 0 x = 2.5 or x = 1 When x = 2.5 y = 7 − 2 x 2.5 = 2      (substituting into A) When x = 1y = 7 − 2 x 1 = 5           (substituting into A) The solutions are (2.5, 2) and (1, 5)

Problems Using Simultaneous Equations

Some word problems can be solved using simultaneous equations.

e.g. Five CDs and nine tapes cost \$204, while six CDs and seven tapes cost \$222.

How much would: (a) One CD cost? (b) One tape cost?

Let one CD cost \$x and one tape cost \$y

From the problem:

5x + 9y = 204 ..................A
6x + 7y = 222 ..................B

Using the elimination method:

Multiply equation A by 6
and multiply equation B by 5

30x + 54y = 1224 ..................C
30x + 35y = 1110 ..................D

Subtract equation D from C

19y = 114
y = 6

Substitute y = 6 into equation A

5x + 9 x 6 = 204
5x = 150
x = 30

(a) The CDs cost \$30 each
(b) The tapes cost \$6 each