Basic Engineering Mathematics, Fifth Edition

74 Basic Engineering Mathematics

Note that subtracting 3 from both sides of the above

equation results in the +3 moving from the LHS to the

RHS, but the sign is changed to –. So, we can move

straight fromx+ 3 =7tox= 7 −3.

Thus, a term can be moved from one side of an equa-

tion to the otheras long as a change in sign is made.

Problem 5. Solve the equation 6x+ 1 = 2 x+ 9

In such equations the terms containingxare grouped

on one side of the equation and the remaining terms

grouped on the other side of the equation. As in Prob-

lems 3 and 4, changing from one side of an equation to

the other must be accompanied by a change of sign.

Since 6 x+ 1 = 2 x+ 9

then 6 x− 2 x= 9 − 1

i.e. 4 x= 8

Dividing both sides by 4 gives

4 x

4

=

8

4

Cancelling gives x= 2

which is the solution of the equation 6x+ 1 = 2 x+9.

In the above examples, the solutions can be checked.

Thus, in Problem 5, where 6x+ 1 = 2 x+9, ifx=2,

then

LHS of equation= 6 ( 2 )+ 1 = 13

RHS of equation= 2 ( 2 )+ 9 = 13

Since the left hand side (LHS) equals the right hand

side (RHS) thenx=2 must be the correct solution of

the equation.

When solving simple equations, always check your

answers by substituting your solution back into the

original equation.

Problem 6. Solve the equation 4− 3 p= 2 p− 11

In order to keep thepterm positive the terms inpare

moved to the RHS and the constant terms to the LHS.

Similar to Problem 5, if 4− 3 p= 2 p− 11

then 4 + 11 = 2 p+ 3 p

i.e. 15 = 5 p

Dividing both sides by 5 gives

15

5

=

5 p

5

Cancelling gives 3 =p or p= 3

which is the solutionof the equation 4− 3 p= 2 p−11.

By substitutingp=3 into the original equation, the

solution may be checked.

LHS= 4 − 3 ( 3 )= 4 − 9 =− 5

RHS= 2 ( 3 )− 11 = 6 − 11 =− 5

Since LHS=RHS, the solutionp=3 must be correct.

If, in this example, the unknown quantities had been

grouped initially on the LHS instead of the RHS, then

− 3 p− 2 p=− 11 − 4

i.e. −^5 p=−^15

from which,

− 5 p

− 5

=

− 15

− 5

and p= 3

as before.

It is often easier, however, to work with positive values

where possible.

Problem 7. Solve the equation 3(x− 2 )= 9

Removing the bracket gives 3x− 6 = 9

Rearranging gives 3 x= 9 + 6

i.e. 3 x= 15

Dividing both sides by 3 gives x= 5

which is the solution of the equation 3(x− 2 )=9.

Theequationmaybecheckedbysubstitutingx=5 back

into the original equation.

Problem 8. Solve the equation

4 ( 2 r− 3 )− 2 (r− 4 )= 3 (r− 3 )− 1

Removing brackets gives

8 r− 12 − 2 r+ 8 = 3 r− 9 − 1

Rearranging gives 8r− 2 r− 3 r=− 9 − 1 + 12 − 8

i.e. 3 r=− 6

Dividing both sides by 3 gives r=

− 6

3

=− 2

which is the solution of the equation

4 ( 2 r− 3 )− 2 (r− 4 )= 3 (r− 3 )− 1.

The solution may be checked by substitutingr=− 2

back into the original equation.

LHS= 4 (− 4 − 3 )− 2 (− 2 − 4 )=− 28 + 12 =− 16

RHS= 3 (− 2 − 3 )− 1 =− 15 − 1 =− 16

Since LHS=RHS thenr=−2 is the correct solution.