Two dozen eggs and a loaf of bread costs K17.50.
Half a dozen eggs and two loaves of bread costs K14.
Find the cost of a dozen egg.
2.0 e + 1 b = 17.50
0.5 e + 2 b = 14.00
4.0 e + 2 b = 35.00
0.5 e + 2 b = 14.00
------------------------- subtract
3.5 e = 21.00
6
Two dozen of eggs and a loaf of a bread cost K8.50.Half a dozen of eggs and two loaves of bread cost K6.50.Find the cost of a dozen eggs
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To find the cost of a dozen eggs, we need to first assign variables to represent the unknown quantities. Let's assume the cost of a dozen eggs is x (in Kina).
We are given two sets of information:
1. Two dozen eggs and a loaf of bread cost K17.50.
This can be represented as: 2x + 1y = 17.50, where y represents the cost of a loaf of bread.
2. Half a dozen eggs and two loaves of bread cost K14.
This can be represented as: 0.5x + 2y = 14.
We now have a system of two equations with two variables:
2x + y = 17.50 (Equation 1)
0.5x + 2y = 14 (Equation 2)
We can solve this system of equations using the method of substitution or elimination. Let's use elimination to eliminate the y variable.
Multiply Equation 2 by 2 to get rid of the decimals:
x + 4y = 28 (Equation 3)
Now, subtract Equation 1 from Equation 3:
(x + 4y) - (2x + y) = 28 - 17.50
x + 4y - 2x - y = 10.50
-x + 3y = 10.50 (Equation 4)
We now have a new equation with only one variable. Let's solve Equation 4 for y:
-x + 3y = 10.50
3y = x + 10.50
y = (x + 10.50)/3 (Equation 5)
Now, substitute the value of y from Equation 5 into Equation 1:
2x + y = 17.50
2x + (x + 10.50)/3 = 17.50
6x + x + 10.50 = 52.50 (multiply both sides by 3 to eliminate the fraction)
7x + 10.50 = 52.50
7x = 52.50 - 10.50
7x = 42
x = 42/7
x = 6
Therefore, the cost of a dozen eggs is K6.