Respuesta :
Answer:
$0.8
Step-by-step explanation:
let x represent the price of apples
let y represent the price of oranges
6x+3y=3.36—eqn 1
2x+5y=3.04—eqn 2
Multiply eqn 2 by 3
3(2x+5y=3.04)=6x+15y=9.12
Subtract eqn 2 from 1
6x+3y=3.36
-{6x+15y=9.12}
-12y=-5.76
y=0.48
plug y=0.48 into eqn 1
6x+3(0.48)=3.36
6x=3.36-1.44
x=1.92/6=0.32
That is, an apple costs $0.32 and an orange costs $0.48
Therefore the cost of an orange and an apple=$0.48+$0.32=$0.8
Answer:
- apple: $0.32
- orange: $0.48
Step-by-step explanation:
Let x and y represent the costs of 1 apple and 1 orange, respectively. The two cost figures tell us ...
6x +3y = 3.36
2x +5y = 3.04
We note that all of the numbers in the first equation are divisible by 3, so we can divide by 3 to reduce it to ...
2x +y = 1.12
Subtracting this from the second equation gives ...
(2x +5y) -(2x +y) = (3.04) -(1.12)
4y = 1.92 . . . . . . . simplify
y = 0.48 . . . . . . . divide by 4
Then the cost of an apple can be found from our reduced equation:
2x +0.48 = 1.12 . . . substitute for y
2x = 0.64 . . . . . . . . subtract 0.48
x = 0.32 . . . . . . . . . divide by 2
The cost of one apple is $0.32; the cost of one orange is $0.48.
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Additional comment
The method of solution we used here is called "elimination," because the x-variable was eliminated from the combined equations. There are numerous other methods available for solving a system of linear equations. Most graphing calculators have features for graphing and for reducing matrices that will solve these for you. All you need to do is enter the equations.
