Respuesta :
Answer:
For Monday
[tex] 55p +22 w = 7.75[/tex] (1)
For Thursday
[tex] 33p +44 w = 10.95[/tex] (2)
[tex] w = 0.20455[/tex]
[tex] p= \frac{7.75-22*0.20455}{55}=0.059091[/tex]
Step-by-step 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
[tex] 55p +22 w = 7.75[/tex] (1)
For Thursday
[tex] 33p +44 w = 10.95[/tex] (2)
And if we solve for p from equation (1) we got:
[tex] p = \frac{7.75-22w}{55}[/tex] (3)
And replacing equation (3) into equation (2) we got:
[tex] 33 (\frac{7.75-22w}{55}) +44w =10.95[/tex]
[tex] 4.65-13.2w +44w = 10.95[/tex]
[tex]30.8w = 10.95-4.65[/tex]
[tex]30.8 w = 6.3[/tex]
[tex] w = 0.20455[/tex]
And the price for the watermellon would be:
[tex] p= \frac{7.75-22*0.20455}{55}=0.059091[/tex]
Answer:
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
Step-by-step 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
