Question

On Monday, Josh went to the Farmer's Market. He bought 55 peaches and 22 watermelons for $7.75.. On Thursday, Josh went back to the Farmer's Market and bought 33 peaches and 44 watermelons for $10.95. What systems of equations can be used to determine the cost for one peach (p)(p) and one watermelon (w)?

Answer 1

The system of equations is

• 5p+2w=7.75

• 3p+4w=10.95.

When solved

• The cost of one watermelon(w) is $2.25
• The cost of one peach(p) is $0.65

Explanation:

Let the price of one peach =p

Let the price of one watermelon=w

On Monday, bought 5 peaches and 2 watermelons for $7.75.

• 5p+2w=7.75

On Thursday, Josh went back to the Farmer's Market and bought 3 peaches and 4 watermelons for $10.95.

• 3p+4w=10.95.

The system of equation that could be used to solve for the price of one peach (p) and one watermelon (w) is therefore:

• 5p+2w=7.75

• 3p+4w=10.95.

To solve this, Multiply the first equation by 4 and the second equation by 2.

• 20p+8w=31
• 6p+8w=21.9

Subtract

14p=9.1

Divide both sides by 14

p=$0.65

Next, we substitute p=$0.65 in any equation to obtain w.

5p+2w=7.75

5(0.65)+2w=7.75

2w=7.75-3.25

2w=4.5

w=$2.25

Therefore:

• The cost of one watermelon(w) is $2.25
• The cost of one peach(p) is $0.65

Answer 2

For Monday

`55p +22 w = 7.75` (1)

For Thursday

`33p +44 w = 10.95` (2)

`w = 0.20455`

`p= (7.75-22*0.20455)/(55)=0.059091`

Explanation:

For this case we define the following notation:

p represent the individual price of peachs

w represent the individual price of water mellons

From the info given we can set up the following equations for the total cost:

For Monday

`55p +22 w = 7.75` (1)

For Thursday

`33p +44 w = 10.95` (2)

And if we solve for p from equation (1) we got:

`p = (7.75-22w)/(55)` (3)

And replacing equation (3) into equation (2) we got:

`33 ((7.75-22w)/(55)) +44w =10.95`

`4.65-13.2w +44w = 10.95`

`30.8w = 10.95-4.65`

`30.8 w = 6.3`

`w = 0.20455`

And the price for the watermellon would be:

`p= (7.75-22*0.20455)/(55)=0.059091`