Six pears and three apples cost $3.90. Two pears and five apples cost $3.30. How much does one pear cost?

Question

help?

Answer 1

im gonna work in cents cuz its ezier

let P be price of pears
let A be price of apples

6P + 3A = 390
3A = 390 - 6P
A = 130 - 2P###Eq.1


2P + 5A = 330###Eq.2


sub Eq.1 into Eq.2
2P + 5 (130-2P) = 330
2P + 650 - 10P = 330
-8P = -320
P = 40 cents

Answer 2

6p + 3a = 3.90
2p + 5a = 3.30

where p = cost of pear, a = cost of apple

Here we have 2 equations and 2 unknowns, so we can solve it!

We want to multiply one equation by a constant, such that when we add the two equations together, one variable will drop out. In the first equation we have 6p, and in the second we have 2p, so if we multiply the second equation by -3, we will have -6p (2p * -3 = -6p) so that when we add 6p and -6p we will get 0, as such:

2p(-3) + 5a(-3) = 3.30(-3)
-6p -15a = -9.90

now when we add the two, we get:

6p + 3a = 3.90
-6p - 15a = -9.90
----------------------
0p -12a = -6

(we add in columns, so add the p's, the a's and the constants)

-12a = -6
-12a/-12 = -6/-12
a = 0.50

so we know an apple is 50 cents, now we can plug it in to either equation to figure out how much pears are:

2p + 5(0.50) = 3.30
2p + 2.50 = 3.30
2p + 2.50 - 2.50 = 3.30 - 2.50
2p = 0.80
p = 0.40

so apples are 50 cents, and pears are 40 cents!

Hope this helps

Answer 3

Using P = pears and A = apples:
6P + 3A = 3.90
2P + 5A = 3.30
We must get down to one variable so multiple the second equation through by a -3 to cancel out the P's. Now we have:
6P + 3A = 3.90
-6P - 15A = -9.90
Now subtract the bottom equation from the top:
0P - 12A = -6.00
Divide through by -12 to isolate and solve for A:
A = 0.50 So apples cost $0.50. Now put 0.50 into one of the equations for A and solve for P. We'll use the first equation:
6P + 3 (0.50) = 3.90
Multiply the 3 by 0.50:
6P + 1.50 = 3.90
Subtract 1.50 from both sides:
6P = 2.40
Divide both sides by 6 to solve for P:
P = 0.40 So pears cost $0.40.
Hope this helps!

Answer 4

Let one pear cost x cents and one apple cost y cents.
According to the problem,
6x+3y=390 .....(eqn no 1) and 2x+5y=330....(eqn. no. 2)
mutilying eqn. no 1 by 5 and eqn.no 2 by 3,we get
30x+15y= 1950
6x+ 15y= 990
(subtracting) 24x=960 =>x=960/24=40
Therefore cost of a pear is 40 cents,

Answer 5

Is there an assumption that the pears and apples don't cost the same?

Answer 6

Pears are .40 cents each.

Don't ask how I came up with it.

Trial & error.

Answer 7

u have to make a system of equations first!

6p+3a=3.90
2p+5a=3.30

Can you solve it from there?