The Chemistry Maths Book, Second Edition

56 Chapter 2Algebraic functions

defines a second function y 1 = 1 g(x). The two equations have simultaneoussolutions

for those values of xfor whichf(x)andg(x) are equal. Graphically, the simultaneous

real solutions are those points, if any, at which the graphs ofy 1 = 1 f(x)andy 1 = 1 g(x)

cross. For example, the two linear equations

a

0

x 1 + 1 b

0

y 1 = 1 c

0

a

1

x 1 + 1 b

1

y 1 = 1 c

1

(2.38)

can be solved to give the solution

(2.39)

We note that this solution exists only if the denominator (a

0

b

1

1 − 1 a

1

b

0

) is not zero.

Graphically, the equations (2.38) then represent two straight lines, and the solution is

the point at which the lines cross.

EXAMPLE 2.31Solve

(1) x+ 1 y 1 = 13

(2) 2x 1 + 13 y 1 = 14

To solve for y, multiply equation (1)by 2 :

(1′)2x 1 + 12 y 1 = 16

(2) 2x 1 + 13 y 1 = 14

and subtract (1′)from (2)to givey 1 = 1 − 2. Substitution for yin (1)then givesx 1 = 15.

The lines therefore cross at point(x,y) 1 = 1 (5, −2)

0 Exercises 66, 67

EXAMPLE 2.32Solve

(1) x 1 + 1 y 1 = 13

(2) 2x 1 + 12 y 1 = 14

To solve, subtract twice (1)from (2):

(1) x 1 + 1 y 1 = 13

(2′)0 1 = 1 − 2

The second equation is not possible. The equations are said to be inconsistent and

there is no solution. This is an example for which the denominator (a

0

b

1

1 − 1 a

1

b

0

)in

equation (2.39) is zero and, graphically, it corresponds to parallel lines.

0 Exercise 68

x

cb cb

ab ab

y

ac ac

ab

=

−

−

,=

−

01 10

01 10

01 10

0

1110

−ab