# The A to Z of mathematics a basic guide - Sidebotham T.H.

ISBN 0-471-15045-2

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Example 1. Nathan and Jacob go into the Cafe Roma and have lunch. Nathan has two sausage rolls and three small muffins at a cost of $3.80. Jacob buys two sausage rolls and two small muffins and his bill is $3.10. From this information find the price of one sausage roll and the price of one small muffin.

Ratio of lengths = |

Then

Cubing the scale factor

1

27

Capacity of the larger bottle = 27 x 10

= 270 ml

SIMULTANEOUS EQUATIONS 411

Solution. Jacob and Nathan buy the same number of sausage rolls, two, but Nathan has one extra muffin. Therefore the difference in what they pay must be the price of one muffin:

Price of one muffin = $3.80 — $3.10 = $0.70

The price of one muffin is 70 cents.

To find the price of a sausage roll we substitute the price of a muffin into either Nathan’s bill or Jacob’s bill:

Cost of 2 sausage rolls + 3 muffins = $3.80

Cost of 2 sausage rolls + 3 x $0.70 = $3.80

Cost of 2 sausage rolls = $3.80 — $2.10 = $1.70

Cost of 1 sausage roll = $0.85

The price of one sausage roll is 85 cents.

Summary: A muffin costs 70 cents and a sausage roll costs 85 cents.

Instead of solving the problem in a descriptive way, we can model the problem with two equations and use the following alternative way of setting out.

Solution. Suppose the price of one sausage roll is r cents, so two will cost Nathan 2r cents, and the price of one muffin is m cents, so three muffins will cost him 3m cents. For Nathan’s order, write

2r + 3m = 380 A

Similarly, for Jacob’s order, Write

2r + 2m = 310 B

We refer to the two equations as A and B.

Subtract the two equations A — B:

2r — 2r + 3m — 2m = 380 — 310 The two equations are subtracted to

eliminate r.

m = 70 2r — 2r = 0, and 3m — 2m = m

This is Nathan’s bill

Substituting $0.70 for the price of one muffin

412 SIMULTANEOUS EQUATIONS

Substituting m = 10 into one of the equations, say A, enables us to find r:

2r + 3 x 70 = 380 2 r + 210 = 380 2 r = 170

r = 85 Dividing both sides of the equation by 2

The two solutions m =10 and r = 85 are the only solutions that solve this problem. We say these solutions are unique.

The following examples explain the various strategies in solving other simultaneous equations.

Example 2. Solve the simultaneous equations

2x + 3y = 3 A

4x — y = 20 B

Solution. The strategy is to eliminate the variable y by multiplying equation B by 3, so that the two coefficients of y are 3 and —3, which are equal except for being opposite in sign:

2x + 3y = 3 A

12x — 3y = 60 C This equation is now called C

Since the coefficients of y in each equation are now both equal, but have opposite signs, we add the two equations to eliminate y:

2x + I2x + 3y — 3y = 3 + 60 A + C

14a: = 63 2a; + 12a: = 14a:, 3y — 3y = 0

x = 4.5 Dividing both sides of the equation by 14

In order to find y, we substitute x = 4.5 in one of the equations, say equation A:

2 x 4.5+ 3y = 3 9 + 3y = 3

3y = —6 Subtracting 9 from both sides of the equation y = -2

The solutions are a; = 4.5, y = —2.

Example 3. Solve the simultaneous equations

2a: + 3y = 4 A

3x +4 y = 6 B

SIMULTANEOUS EQUATIONS 413

Solution. The strategy is to eliminate the variable y by multiplying equation A by 4 and equation B by 3, so that the coefficients of y in both equations are the same, which is equal to 12:

%x + 12 y = 16 C

9x + 12y = 18 D The equations are renamed C and D

Since the coefficients of y in each equation are now both the same and have the same signs, we subtract the two equations to eliminate y:

8jc-9x + 12y-12y = 16-18 C-D —x = —2

x = 2 Multiplying both sides of the equation by — 1

In order to find y we substitute x = 2 in one of the equations, say equation A:

2 x 2 + 3y = 4 4 + 3y = 4

3y = 0 Subtracting 4 from both sides of the equation

y = 0 Dividing both sides of the equation by 3.

The solutions are x = 2, y = 0.

Some questions are more easily solved by substitution instead of the “elimination” method described above. The substitution method is explained here.

Example 4. Solve the simultaneous equations

y = —2x + 5 A 3y — 2x = 47 B

Solution. One of the equations is written as y = ..., so the strategy is to substitute (—2x + 5) for y in equation B; it is advised to use brackets around —2x + 5. Write

3(-2x + 5) - 2x = 47 Substituting for y in equation B

—6x + 15 — 2x = 47 Expanding the brackets

—8x + 15 = 47 —6x —2x — — 8x

—8x = 32 Subtracting 15 from both sides of the equation

x = -4 Dividing both sides of the equation by —8

414 SIMULTANEOUS EQUATIONS

To find y we substitute x — —4 into equation A:

y = -2 x -4 + 5 y = 8 + 5 > = 13

The solutions arex = —4, y = 13.

So far we have used the “elimination” method and the “substitution” method for solving simultaneous equations. Another method is to draw straight-line graphs of the two equations and read off from the graphs the coordinates of the point where the two straight lines intersect. Naturally if the answers involve decimals, this method may not give perfectly accurate results. Before continuing with this explanation, make sure you know how to draw linear graphs, which treated is under the entry Gradient-Intercept Form.

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