Q. What Are Systems of Simultaneous Linear Equations?
A system of simultaneous linear equations is a collection of two or more linear equations involving multiple variables, where the equations are solved together because the variables are interdependent.
Each equation represents a relationship between variables, and the solution is a set of values that satisfies all equations simultaneously.
Its uses in Economics
In economics, many variables are interrelated. For instance:
• Consumption depends on income,
• Income depends on investment,
• Investment may depend on interest rates.
These relationships are modelled using simultaneous equations to understand how changes in one variable affect others within an economic system.
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NP1: Solve the given system of simultaneous equations:
2x + 3y = 12
x + y = 5
Solution:
From (2):
x = 5 − y
Substitute into (1):
2(5 − y) + 3y = 12
⇒10 − 2y + 3y = 12
⇒ y = 22
Now plug y = 2 into (2):
x + 2 = 5
⇒ x = 3
∴ x = 3, y = 2
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NP2: Solve the given system of simultaneous equations:
x + 3y = 7 −−−−(1)
2x − y = 4 −−−−(2)
Solution:
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Application in Economics:
NP1: A commodity is produced by 3 units of labour and 2 units of capital. The total cost comes to 62. If the commodity is produced by using 4 units of labour and 1 unit of capital, the cost comes to 56. What is the cost per unit of labour and capital?
Solution:
Given:
3L+2K=62−−−−(1)
4L+1K=56 −−−−(2)
Where:
• L is the cost per unit of labour
• K is the cost per unit of capital
Solving by elimination method:
To do that, multiply Equation 2 by 2:
2(4L + K)= 2(56)
⇒8L + 2K = 112−−−−(3)
Subtract (1) from (3):
(8L + 2K)−(3L + 2K)=112−62
(8L − 3L)+(2K − 2K)=50
⇒5L=50
⇒L=10
Substitute L=10 in (2)
4L + K = 56
⇒ 4(10) + K = 56
⇒ 40 + K = 56
⇒ K = 56 − 40
⇒ K = 16
∴ Cost per unit of Labour (L) = ₹10 and cost per unit of Capital (K) = ₹16
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NP2: A product is made using 5 units of labour and 2 units of capital, costing ₹74. Another product uses 3 units of labour and 4 units of capital, costing ₹78. Find the cost per unit of labour and capital.
Solution:
Given:
5L+2K=74−−−−(1)
3L+4K=78 −−−−(2)
Where:
L is the cost per unit of labour
K is the cost per unit of capital
Solving by elimination method:
To do that, multiply Equation (1) by 2:
2(5L + 2K) = 2(74)
⇒10L + 4K = 148−−−−(3)
Subtract (2) from (3):
(10L + 4K) − (3L + 4K) = 148 − 78
10L + 4K − 3L − 4K = 70
7L = 70
⇒L = 10
Substitute L=10 in (1)
5L + 2K = 74
⇒5(10) + 2K = 74
⇒50 + 2K = 74
⇒2K = 24
⇒K = 12
∴ Cost per unit of Labour (L) = ₹10 and cost per unit of Capital (K) = ₹12
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NP3: A consumer buys two goods: Good A and Good B. When they buy 4 units of A and 6 units of B, they spend ₹72. When they buy 6 units of A and 3 units of B, they spend ₹75. Find the price per unit of Good A and Good B.
Solution:
Given:
Let:
A = price per unit of Good A
B = price per unit of Good B
4A + 6B = 72−−−−(1)
6A + 3B = 75−−−−(2)
Solving by elimination method:
To do that, multiply Equation (2) by 2:
2(6A + 3B) = 2(75)
⇒12A + 6B = 150−−−−(3)
Subtract (2) from (3):
(12A + 6B) − (4A + 6B)=150−72
12A + 6B − 4A − 6B = 150 − 72
8A = 78
∴ The price per unit of Good A = 9.75 and Good B = 5.5
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NP4: A producer uses 2 tonnes of steel and 3 tonnes of copper to make a product costing ₹132. Another variation uses 5 tonnes of steel and 1 tonne of copper, costing ₹140. Find price per tonne of steel and copper.
Solution:
Given:
Let:
S = price of steel
C = price of copper
2S + 3C = 132−−−−(1)
5S + 1C = 140−−−−(2)
Solving by elimination method:
To do that, multiply Equation (2) by 3:
3(5S + 1C) = 3 (140)
⇒15S + 3C = 420−−−−(3)
Subtract (1) from (3):
(15S + 3C) − (2S + 3C) = 420−132
15S + 3C − 2S − 3C = 288
13S = 288
S = 288/13
S = 22.15
Substitute S=22.15 in (1)
5(22.15) + C = 140
⇒110.75+C = 140
⇒ C = 140 − 110.75
⇒ C = 140 − 110.75
⇒ C = 29.25
∴The price per tonne of steel and copper are:
Steel = ₹22.15/tonne
Copper = ₹29.25/tonne
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