# If 5 apples and 4 oranges cost $ 3.40 while 7 apples and 6 oranges cost $ 4.90, find the cost of an apple and an orange.

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We'll establish the cost of an apple as x and the cost of an orange as y.

We'll write mathematically the phrase "5 apples and 4 oranges cost $ 3.40":

5x + 4y = 3.4 (1)

We'll write mathematically the phrase " 7 apples and 6 oranges cost $ 4.90":

7x + 6y = 4.9 (2)

We'll solve the system using elimination method. We'll add 6*eq.(1) + (-4)*eq.(2):

30x + 24y - 28x - 24y = 20.4 - 19.6

We'll combine and eliminate like terms:

2x = 0.8

x = 0.8/2

x = $ 0.4

We'll substitute x in eq.(1):

5*0.4 + 4y = 3.4

2 + 4y = 3.4

4y = 3.4 - 2

4y = 1.4

y = 1.4/4

y = 0.7/2

y = $ 0.35

So, the cost of an apple is $ 0.4 and the cost of an orange is $ 0.35.

Let the price of an apple be x and an orange be y.

Then the cost of 5 apples and 4 oranges together is 5x+4y which is given to be $3.4.

The cost of 7apples and 6 oranges together is 7x+6y which is given to be $4.9.

Therefore the required equations are:

5x+4y = 3.4.....(1).

7x+6y = 4.9.....(2).

(1)*3-(2)*2 gives: 15x-14x = 3.4*3-4.9*2 = 0.4

Therefore x = 0.4.

Put x = 0.4 in (1): 5*0.4 +4y = 3.4.

Therefore 2 +4y = 1.4.

Therefore an apple cost $0.4 and an orange cost $0.35.

4y = 3.4-2 = 1.4.

y = 1.4/4 = 0.35