basic_engineering_mathematics0

48 Basic Engineering Mathematics
one side of an equation to the other as long as a change in sign
is made.
Problem 5. Solve 6x + 1 = 2x + 9
In such equations the terms containing x are grouped on one side
of the equation and the remaining terms grouped on the other
side of the equation. As in Problems 3 and 4, changing from one
side of an equation to the other must be accompanied by a change
of sign.
Thus since 6x + 1 = 2x + 9
then 6x − 2x = 9 − 1
4x = 8
4x
4 = 8 4
i.e. x = 2
Check: LHS of original equation = 6(2) + 1 = 13
RHS of original equation = 2(2) + 9 = 13
Hence the solution x = 2 is correct.
Problem 6. Solve 4 − 3p = 2p − 11
In order to keep the p term positive the terms in p are moved to
the RHS and the constant terms to the LHS.
Removing the bracket gives: 3x − 6 = 9
Rearranging gives: 3x = 9 + 6
3x = 15
3x
3 = 15 3
i.e. x = 5
Check:
LHS = 3(5 − 2) = 3(3) = 9 = RHS
Hence the solution x = 5 is correct.
Problem 8. Solve 4(2r − 3) − 2(r − 4) = 3(r − 3) − 1
Removing brackets gives:
8r − 12 − 2r + 8 = 3r − 9 − 1
Rearranging gives: 8r − 2r − 3r =−9 − 1 + 12 − 8
i.e. 3r =−6
r = −6
3 = −2
Check: LHS = 4(−4 − 3) − 2(−2 − 4) =−28 + 12 =−16
RHS = 3(−2 − 3) − 1 =−15 − 1 =−16
Hence the solution r =−2 is correct.
Hence
4 + 11 = 2p + 3p
Now try the following exercise
15 = 5p
15
5 = 5p
5
Hence 3 = p or p = 3
Check: LHS = 4 − 3(3) = 4 − 9 =−5
RHS = 2(3) − 11 = 6 − 11 =−5
Hence the solution p = 3 is correct.
If, in this example, the unknown quantities had been grouped
initially on the LHS instead of the RHS then:
−3p − 2p =−11 − 4
i.e. −5p =−15
−5p
−5 = −15
−5
and p = 3, as before
It is often easier, however, to work with positive values where
possible.
Probelm 7. Solve 3(x − 2) = 9
Exercise 26
Further problems on simple equations
(Answers on page 273)
Solve the following equations:
1. 2x + 5 = 7 2. 8− 3t = 2
2
3. c − 1 = 3 4. 2x −1 = 5x + 11
3
5. 7 − 4p = 2p − 3 6. 2.6x − 1.3 = 0.9x + 0.4
7. 2a + 6 − 5a = 0 8. 3x − 2 − 5x = 2x − 4
9. 20d − 3 + 3d = 11d + 5 − 8
10. 2(x − 1) = 4
11. 16 = 4(t + 2)
12. 5( f − 2) − 3(2f + 5) + 15 = 0
13. 2x = 4(x − 3)
14. 6(2 − 3y) − 42 =−2(y − 1)
15. 2(3g − 5) − 5 = 0
16. 4(3x + 1) = 7(x + 4) − 2(x + 5)
17. 10 + 3(r − 7) = 16 − (r + 2)
18. 8 + 4(x − 1) − 5(x − 3) = 2(5 − 2x)