Cambridge International Mathematics

Algebra (Equations and inequalities) (Chapter 3) 81

Example 8 Self Tutor

Solve forx:

2 x+3

4

=

x¡ 2

3

2 x+3

4

=

x¡ 2

3

fLCD=12g

)

3 £( 2 x+3)

3 £ 4

=

4 £(x¡ 2 )

4 £ 3

fto achieve a common denominatorg

) 3(2x+3)=4(x¡2) fequating numeratorsg

) 6 x+9=4x¡ 8 fexpanding bracketsg

) 6 x+9¡ 4 x=4x¡ 8 ¡ 4 x fsubtracting 4 xfrom both sidesg

) 2 x+9=¡ 8

) 2 x+9¡ 9 =¡ 8 ¡ 9 fsubtracting 9 from both sidesg

) 2 x=¡ 17

) fdividing both sides by 2 g

2 x

2

=¡

17

2

) x=¡ (^812)
Example 9 Self Tutor
Solve forx:
x
3

¡

1 ¡ 2 x

6

=¡ 4

x

3

¡

1 ¡ 2 x

6

=¡ 4 fLCD=6g

)

x

3

£

2

2

¡

μ

1 ¡ 2 x

6

¶

=¡ 4 £

6

6

fto create a common denominatorg

) 2 x¡(1¡ 2 x)=¡ 24 fequating numeratorsg

) 2 x¡1+2x=¡ 24 fexpandingg

) 4 x¡1=¡ 24

) 4 x¡ 1 +1=¡ 24 +1 fadding 1 to both sidesg

) 4 x=¡ 23

) x=¡^234 fdividing both sides by 4 g

UNKNOWN IN THE DENOMINATOR

If the unknown appears as part of the denominator, we still solve by:

² writing the equations with thelowest common denominator (LCD)and then

² equating numerators.

IGCSE01

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Y:\HAESE\IGCSE01\IG01_03\081IGCSE01_03.CDR Monday, 15 September 2008 10:40:07 AM PETER