Basic Engineering Mathematics, Fifth Edition

13 Solving simultaneous equations

Now try the following Practice Exercise

PracticeExercise 42 Solving simple

equations (answers on page 344)

Solve the following equations.

1. 2x+ 5 = 7


2. 8− 3 t= 2

3.

2

3

c− 1 = 3

4. 2x− 1 = 5 x+ 11


5. 7− 4 p= 2 p− 5


6. 
   
   
   
   2. 6 x− 1. 3 = 0. 9 x+ 0. 4
   
   
   
   


7. 2a+ 6 − 5 a= 0


8. 3x− 2 − 5 x= 2 x− 4


9. 20d− 3 + 3 d= 11 d+ 5 − 8


10. 2(x− 1 )= 4


11. 16= 4 (t+ 2 )


12. 5(f− 2 )− 3 ( 2 f+ 5 )+ 15 = 0


13. 2x= 4 (x− 3 )


14. 6( 2 − 3 y)− 42 =− 2 (y− 1 )


15. 2( 3 g− 5 )− 5 = 0


16. 4( 3 x+ 1 )= 7 (x+ 4 )− 2 (x+ 5 )


17. 11+ 3 (r− 7 )= 16 −(r+ 2 )


18. 8+ 4 (x− 1 )− 5 (x− 3 )= 2 ( 5 − 2 x)

Here are some further worked examples on solving
simple equations.

Problem 9. Solve the equation

4

x

=

2

5

The lowest common multiple (LCM) of the denomina-
tors, i.e. the lowest algebraic expression that bothxand
5 will divide into, is 5x.

Multiplying both sides by 5xgives

5 x

(

4

x

)

= 5 x

(

2

5

)

Cancelling gives 5( 4 )=x( 2 )

i.e.^20 =^2 x (1)

Dividing both sides by 2 gives

20

2

=

2 x

2

Cancelling gives 10 =xorx= 10

which is the solution of the equation

4

x

=

2

5

When there is just one fraction on each side of the equa-

tion as in this example, there is a quick way to arrive

at equation (1) without needing to find the LCM of the

denominators.

We can move from

4

x

=

2

5

to 4× 5 = 2 ×xby what is

called ‘cross-multiplication’.

In general, if

a

b

=

c

d

thenad=bc

We can use cross-multiplication when there is one

fraction only on each side of the equation.

Problem 10. Solve the equation

3

t− 2

=

4

3 t+ 4

Cross-multiplication gives^3 (^3 t+^4 )=^4 (t−^2 )

Removing brackets gives 9t+ 12 = 4 t− 8

Rearranging gives 9 t− 4 t=− 8 − 12

i.e. 5 t=− 20

Dividing both sides by 5 gives t=

− 20

5

=− 4

which is the solution of the equation

3

t− 2

=

4

3 t+ 4

Problem 11. Solve the equation

2 y

5

+

3

4

+ 5 =

1

20

−

3 y

2

The lowest common multiple (LCM) of the denomina-

tors is 20; i.e., the lowest number that 4, 5, 20 and 2 will

divide into.

Multiplying each term by 20 gives

20

(

2 y

5

)

+ 20

(

3

4

)

+ 20 ( 5 )= 20

(

1

20

)

− 20

(

3 y

2

)

Cancelling gives 4 ( 2 y)+ 5 ( 3 )+ 100 = 1 − 10 ( 3 y)

i.e. 8 y+ 15 + 100 = 1 − 30 y

Rearranging gives 8 y+ 30 y= 1 − 15 − 100