basic_engineering_mathematics0

62 Basic Engineering Mathematics
Now try the following exercise
Exercise 33
Further problems on simultaneous
equations (Answers on page 274)
Solve the following simultaneous equations and verify the
results.
1. a + b = 7 2. 2x + 5y = 7
a − b = 3 x + 3y = 4
3. 3s + 2t = 12 4. 3x − 2y = 13
4s − t = 5 2x + 5y =−4
5. 5m − 3n = 11 6. 8a − 3b = 51
3m + n = 8 3a + 4b = 14
7. 5x = 2y 8. 5c = 1 − 3d
3x + 7y = 41 2d + c + 4 = 0
9.3 Further worked problems on
simultaneous equations
Problem 5.
Solve
3p = 2q (1)
4p + q + 11 = 0 (2)
Rearranging gives:
3p − 2q = 0 (3)
4p + q =−11 (4)
Multiplying equation (4) by 2 gives:
8p + 2q =−22 (5)
Adding equations (3) and (5) gives:
11p + 0 =−22
p = −22
11 =−2
Substituting p =−2 into equation (1) gives:
3(−2) = 2q
−6 = 2q
q = −6
2 =−3
Checking, by substituting p =−2 and q =−3 into equation (2)
gives:
LHS = 4(−2) + (−3) + 11 =−8 − 3 + 11 = 0 = RHS
Hence the solution is p =−2, q =−3
Problem 6.
Solve
x
8 + 5 2 = y (1)
13 − y = 3x
3
(2)
Whenever fractions are involved in simultaneous equations it is
usual to firstly remove them. Thus, multiplying equation (1) by
8 gives:
( x
)
8 + 8
8
( 5
2
)
= 8y
i.e. x + 20 = 8y (3)
Multiplying equation (2) by 3 gives:
39 − y = 9x (4)
Rearranging equations (3) and (4) gives:
x − 8y =−20 (5)
9x + y = 39 (6)
Multiplying equation (6) by 8 gives:
72x + 8y = 312 (7)
Adding equations (5) and (7) gives:
73x + 0 = 292
x = 292
73 = 4
Substituting x = 4 into equation (5) gives:
4 − 8y =−20
4 + 20 = 8y
24 = 8y
y = 24
8 = 3
Checking: substituting x = 4, y = 3 in the original equations,
gives:
Equation (1) :
LHS = 4 8 + 5 2 = 1 2 + 2 1 2 = 3 = y = RHS
Equation (2) : LHS = 13 − 3 = 13 − 1 = 12
3
RHS = 3x = 3(4) = 12
Hence the solution is x = 4, y = 3
Problem 7. Solve
2.5x + 0.75 − 3y = 0
1.6x = 1.08 − 1.2y