Respuesta :
Hey there
Question no.1:
Debbie bought 8 oranges and 7 apples for $10.20. Katrina bought 5 oranges and 14 apples for $12.15. How much did one apple and one orange cost?
Solution:
Let the cost of one apple be "y" and the cost of one orange be "x".
Here,
According to question,
8x + 7y = $10.20 5x + 14y = $12.15
=> 5(8x + 7y) = 10.20 × 5 => 8(5x + 14y) = 12.15*8
=> 40x + 35y = 51 -->(i) => 40x + 112y = 97.2-->(ii)
By Elimination method,
Equation (i) - (ii)
(40x + 35y) - (40x + 112y) = 51 - 97.2
=> 40x + 35y - 40x - 112y = - 46.2
=> - 77y = - 46.2
=> y = 46.2/77
=> y = 0.6
Putting the value of "y" in equation (i) we get,
40x + 35y = 51
=> 40x + 35 × 0.6 = 51
=> 40x + 21 = 51
=> 40x = 51 - 21
=> 40x = 30
=> x = 0.75
Therefore the cost of one orange is $0.75 and the cost of one apple is $ 0.6.
Question no.2:
Joel and Janie are selling fruit. Joel sold 2 small boxes of fruit and 5 large boxes for $31. Janie sold 4 small boxes of fruit and 7 large boxes for $47. How much does one small box of fruit and one large box of fruit cost?
Solution:
Let the cost of a small box be "x" and the cost of a large box be "y".
Here,
According to question,
2x + 5y = $31 4x + 7y = $47
4(2x + 5y) = 31 × 4 2(4x + 7y) = 47 × 2
=> 8x + 20y = 124 ->(i) => 8x + 14y = 94 ->(ii)
By Elimination method,
Equation (i) - (ii)
(8x + 20y) - (8x + 14y) = 124 - 94
=> 8x + 20y - 8x - 14y = 30
=> 6y = 30
=> y = 30/6
=> y = 5
Putting the value of "y" in Equation (ii) we get,
8x + 14y = 94
=> 8x + 14 × 5 = 94
=> 8x + 70 = 94
=> 8x = 94 - 70
=> 8x = 24
=> x = 24/8
=> x = 3