Download simultaneous linear equations

Simultaneous Linear
Equations
15
Introduction
In previous classes, we have read that an equation of the form ax + b = 0, where a, b are real
numbers and a ≠ 0, is called a linear equation in one variable. We have also learnt that every
linear equation in one variable has a unique solution.
In this chapter, we shall extend our knowledge of linear equations in one variable to linear
equations in two variables and we shall also learn the various methods of solving a pair or a
system of two linear equations in two variables.
5.1 Simultaneous linear equations
Linear equation in two variables
An equation of the form ax + by + c = 0, where a, b and c are real numbers and a and b are non-zero, is
called a general linear equation in the two variables x and y.
For example, x + y – 3 = 0 is a linear equation in the two variables (unknowns) x and y.
Solution of a linear equation in two variables
x = α and y = β is a solution of the linear equation ax + by + c = 0 if and only if
aα + bβ + c = 0, where α, β are real numbers.
Every linear equation in two variables has an unlimited number of solutions.
For example, x = 0, y = 3 ; x = 1, y = 2 ; x = 2, y = 1 ; x = 3, y = 0 and x = 7,
y = – 4 etc. are all solutions of the equation x + y – 3 = 0.
System of simultaneous linear equations
Let us consider two linear equations in two variables,
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0.
These two equations are said to form a system of simultaneous linear equations.
For example,
x + y – 3 = 0
2x – 5y + 1 = 0
is a system of two simultaneous linear equations in the two variables x and y.
A solution to a system of two simultaneous linear equations in two variables is an ordered pair of
numbers which satisfy both the equations.
For the above example, x = 2, y = 1 is a solution to the system of simultaneous linear
equations. We can check this by substituting x = 2, y = 1 into each of these two equations.
If there is only one such solution, then the system of linear equations is said to be consistent
and independent. In this book, we shall be dealing only with such a system of simultaneous
linear equations.
The various methods of solving a pair or a system of linear equations are:
(i)Substitution method.
(ii)Elimination method.
(iii) Cross-multiplication method.
We shall discuss these methods one by one.
5.2 Substitution Method
Procedure:
(i)Solve one of the given equations for one of the variables, whichever is convenient.
(ii)Substitute that value of the variable in the other equation.
(iii)Solve the resulting single variable equation. Substitute this value into either of the
two original equations, and solve it to find the value of the second variable.
Remark
The solution may be checked by substituting in both the original equations.
Illustrative Examples
Example 1. Solve the following system of linear equations:
4x – 3y = 8
x – 2y = – 3.
Solution. The given equations are
4x – 3y = 8
…(i)
x – 2y = – 3
…(ii)
We can solve either equation for either variable. But to avoid fractions, we solve the second
equation for x,
x = 2y – 3
Substituting this value of x in equation (i), we get
4(2y – 3) – 3y = 8
⇒ 8y – 12 – 3y = 8
⇒ 5y = 20 ⇒ y = 4.
Substituting this value of y in (ii), we get
x – 2 × 4 = – 3 ⇒ x – 8 = – 3 ⇒ x = 5.
Hence, the solution is x = 5, y = 4.
Example 2. Solve the following system of linear equations:
8x + 5y = 9
3x + 2y = 4.
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…(iii)
Solution. The given system of simultaneous linear equations is
8x + 5y = 9
…(i)
3x + 2y = 4
…(ii)
From equation (ii), we get
2y = 4 – 3x ⇒ y = 4 − 3 x .
2
Substituting this value of y in equation (i), we get
8x + 5 . 4 − 3 x = 9
2
⇒ 16x + 20 – 15x = 18 ⇒ x + 20 = 18
⇒ x = – 2.
Substituting this value of x in equation (ii), we get
3 × (– 2) + 2y = 4
⇒ – 6 + 2y = 4 ⇒ 2y = 10 ⇒ y = 5.
Hence, the solution is x = – 2, y = 5.
Example 3. Solve the following pair of linear equations:
3x 5 y
−
= – 2
2
3
x y
13
=
.
+
3 2
6
Solution. The given equations are:
3x 5y
−
= – 2
2
3
…(i)
x y 13
+ =
3 2
6
and
…(ii)
Multiplying both sides of the equations (i) and (ii) by 6, we get
9x – 10y = –12
…(iii)
From equation (iv), we get y =
and
2x + 3y = 13
13 − 2 x
3
…(iv)
…(v)
Substituting this value of y in equation (iii), we get
13 − 2 x 
9x – 10 
 = –12

3

⇒ 27x – 130 + 20x = – 36
(Multiplying both sides by 3)
⇒ 47x = 94 ⇒ x = 2.
Substituting this value of x in equation (v), we get
y =
13 − 2 × 2 13 − 4 9
=
= = 3.
3
3
3
Hence, the solution is x = 2 and y = 3.
Example 4. Solve 2x + 3y = 11 and 2x – 4y = – 24. Hence, find the value of ‘m’ for which
y = mx + 3.
Solution. The given equations are
2x + 3y = 11
…(i)
and
2x – 4y = – 24
4 y − 24
From equation (ii), we get x =
2
⇒ x = 2y – 12
Simultaneous linear equations
…(ii)
…(iii)
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Substituting this value of x in equation (i), we get
2(2y – 12) + 3y = 11
⇒ 4y + 3y – 24 = 11 ⇒ 7y = 35 ⇒ y = 5.
Substituting this value of y in equation (iii), we get
x = 2 × 5 – 12 = 10 – 12 = – 2.
Hence, the solution is x = – 2 and y = 5.
Now, y = mx + 3
…(iv)
Putting x = – 2 and y = 5 in (iv), we get
5 = m(– 2) + 3 ⇒ 2m = 3 – 5
⇒ 2m = – 2 ⇒ m = – 1.
Hence, the value of m is – 1.
Exercise 5.1
Solve the following systems of simultaneous linear equations by the substitution method (1 to 4):
1. (i) x + y = 14 (ii) s – t = 3
x – y = 4
s t
+ =6
3 2
(iii) 2x + 3y = 9 (iv) 3x – 5y = 4
3x + 4y = 5 9x – 2y = 7.
2. (i) a + 3b = 5
(ii) 5x + 4y – 4 = 0
7a – 8b = 6 x – 20 = 12y.
3. (i) 2x – 3 y = 3
4
(ii) 2x + 3y = 23
5x – 2y – 7 = 0 5x – 20 = 8y.
4. (i) mx – ny = m2 + n2
(ii)
b
a
x + y = a2 + b2
a
b
x + y = 2m x + y = 2ab.
x
5. Solve 2x + y = 35, 3x + 4y = 65. Hence, find the value of y .
6. Solve the simultaneous equations 3x – y = 5, 4x – 3y = – 1. Hence, find p, if y = px – 3.
5.3 elimination method
ow let us consider another method of eliminating (removing) one variable. This method is
N
sometimes more convenient than substitution method.
Procedure:
(i)Multiply one or both equations (if necessary) by a suitable number(s) to transform
them so that addition or subtraction will eliminate one variable.
(ii)Solve the resulting single variable equation and substitute this value into either of the
two original equations, and solve it to find the value of the second variable.
Remark. I n particular, if the coefficient of x in the first equation is numerically equal to the
coefficient of y in the second equation and the coefficient of y in the first equation
is numerically equal to the coefficient of x in the second equation, then add and
subtract the given equations. This gives the values of x + y and x – y. Then find
the values of x and y. (See example 3)
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