96 Basic Engineering Mathematics
Substituting x = 2 into equation (3) gives
5(2) + 3y = 1
10 + 3y = 1
3y = 1 − 10 = −9
y =
−9
3
= −3
Checking, substituting x = 2, y = −3 in equation (4)
gives
LHS = −2 + 3(−3) = −2 − 9 = −11 = RHS
Hence, the solution is x = 2, y = −3.
Now try the following Practice Exercise
Practice Exercise 51 Solving more difficult
simultaneous equations (answers on page
345)
In problems 1 to 7, solve the simultaneous equations and verify the results
1.
3
x
+
2
y
= 14
2.
4
a
−
3
b
= 18
5
x
−
3
y
= −2
2
a
+
5
b
= −4
3.
1
2 p
+
3
5q
= 5
4 .
5
x
+
3
y
= 1.1
5
p
−
1
2q
=
35
2
3
x
−
7
y
= −1.1
5.
c + 1
4
−
d + 2
3
+ 1 = 0
1 − c
5
+
3 − d
4
+
13
20
= 0
7.
5
x + y
=
20
27
4
2x − y
=
16
33
6.
3r + 2
5
−
2s − 1
4
=
11
5
3 + 2r
4
+
5 − s
3
=
15
4
8. If 5x −
3
y
= 1 and x +
4
y
=
5
2
, find the value
of
x y + 1
y
13.5 Practical problems involving
simultaneous equations
There are a number of situations in engineering and
science in which the solution of simultaneous equations
is required. Some are demonstrated in the following
worked problems.
Problem 12. The law connecting friction F and
load L for an experiment is of the form F = aL + b
where a and b are constants. When F = 5.6 N,
L = 8.0 N and when F = 4.4N , L = 2.0 N. Find the
values of a and b and the value of F when
L = 6.5 N
Substituting F = 5.6 and L = 8.0 into F = aL + b
gives
5.6 = 8.0a + b
(1)
Substituting F = 4.4 and L = 2.0 into F = aL + b
gives
4.4 = 2.0a + b
(2)
Subtracting equation (2) from equation (1) gives
1.2 = 6.0a
a =
1.2
6.0
=
1
5
or 0.2
Substituting a =
1
5
into equation (1) gives
5.6 = 8.0
1
5
+ b
5.6 = 1.6 + b
5.6 − 1.6 = b
i.e.
b = 4
Checking, substituting a =
1
5
and b = 4 in equation (2),
gives
RHS = 2.0
1
5
+ 4 = 0.4 + 4 = 4.4 = LHS
Hence, a =
1
5
and b = 4
When L = 6.5, F = aL + b =
1
5
(6.5) + 4 = 1.3 + 4,
i.e. F = 5.30 N.
Substituting x = 2 into equation (3) gives
5(2) + 3y = 1
10 + 3y = 1
3y = 1 − 10 = −9
y =
−9
3
= −3
Checking, substituting x = 2, y = −3 in equation (4)
gives
LHS = −2 + 3(−3) = −2 − 9 = −11 = RHS
Hence, the solution is x = 2, y = −3.
Now try the following Practice Exercise
Practice Exercise 51 Solving more difficult
simultaneous equations (answers on page
345)
In problems 1 to 7, solve the simultaneous equations and verify the results
1.
3
x
+
2
y
= 14
2.
4
a
−
3
b
= 18
5
x
−
3
y
= −2
2
a
+
5
b
= −4
3.
1
2 p
+
3
5q
= 5
4 .
5
x
+
3
y
= 1.1
5
p
−
1
2q
=
35
2
3
x
−
7
y
= −1.1
5.
c + 1
4
−
d + 2
3
+ 1 = 0
1 − c
5
+
3 − d
4
+
13
20
= 0
7.
5
x + y
=
20
27
4
2x − y
=
16
33
6.
3r + 2
5
−
2s − 1
4
=
11
5
3 + 2r
4
+
5 − s
3
=
15
4
8. If 5x −
3
y
= 1 and x +
4
y
=
5
2
, find the value
of
x y + 1
y
13.5 Practical problems involving
simultaneous equations
There are a number of situations in engineering and
science in which the solution of simultaneous equations
is required. Some are demonstrated in the following
worked problems.
Problem 12. The law connecting friction F and
load L for an experiment is of the form F = aL + b
where a and b are constants. When F = 5.6 N,
L = 8.0 N and when F = 4.4N , L = 2.0 N. Find the
values of a and b and the value of F when
L = 6.5 N
Substituting F = 5.6 and L = 8.0 into F = aL + b
gives
5.6 = 8.0a + b
(1)
Substituting F = 4.4 and L = 2.0 into F = aL + b
gives
4.4 = 2.0a + b
(2)
Subtracting equation (2) from equation (1) gives
1.2 = 6.0a
a =
1.2
6.0
=
1
5
or 0.2
Substituting a =
1
5
into equation (1) gives
5.6 = 8.0
1
5
+ b
5.6 = 1.6 + b
5.6 − 1.6 = b
i.e.
b = 4
Checking, substituting a =
1
5
and b = 4 in equation (2),
gives
RHS = 2.0
1
5
+ 4 = 0.4 + 4 = 4.4 = LHS
Hence, a =
1
5
and b = 4
When L = 6.5, F = aL + b =
1
5
(6.5) + 4 = 1.3 + 4,
i.e. F = 5.30 N.