Lydia wants to buy picnic tables, benches, and permanently installed grills. On one page she notices that 5 picnic tables, 10 benches, and 3 grills sell for $1330. On another page she sees that 20 picnic tables, 30 benches, and 20 grills sell for $5550. Finally, on a third page she sees that 2 picnic tables and 4 benches sell for $430. Help Lydia find the individual prices for the three items. Define three variables, write a system of equations, and solve to find the cost of each item. Justify your answer by showing how you solved the problem, and check the solution.
5p+10b+3g= 1330-->-4(5p+10b+3g)=-4(1330)
-20p-40b-12g= -5320
20p+30b+20g= 5550
---------------------------
-10b+8g= 230
5p+10b+3g=1330
2p+4b+ 0g?= 430
im not really sure how to set up this problem...please help!
looks like you have 3 equations:
5p+10b+3g = 1330 #1
2p+3b+2g = 555 #2 (I divided all terms by 10)
p+2b = 215 #3
from #2 p = 215-2b
sub into #1 5(215-2b)+10b+3g = 1330
---> 1075 - 10b + 10b + 3g = 1330
g = 85 , that was lucky, the 10b dropped out
back into #2
2(215-2b) + 3b + 170 = 555
solving this, gave me b = 45
now into #3
p +90 = 215
p = 125
Picnic table costs $125
Bench costs $45 and
Grill costs $85
omg thanks SO much! :)
To solve this problem, we need to set up a system of equations using the given information.
Let's define the variables:
p = price of each picnic table
b = price of each bench
g = price of each grill
From the first page, we can write the equation:
5p + 10b + 3g = 1330
From the second page, we can write the equation:
20p + 30b + 20g = 5550
And from the third page, we can write the equation:
2p + 4b = 430
Now, we have a system of three equations with three variables. To solve this system, we can use the method of substitution or elimination.
Let's use elimination to solve the system. Multiply the first equation by 4, and the third equation by 5 (to get the same coefficient for 'b'):
20p + 40b + 12g = 5320
10p + 20b = 2150
Subtract the second equation from the first:
(20p + 30b + 20g) - (10p + 20b) = 5550 - 2150
10p + 10b + 20g = 3400
Divide the equation by 10:
p + b + 2g = 340
Now we have the following equations:
p + b + 2g = 340 -------- Equation 1
2p + 4b = 430 -------- Equation 2
We can solve Equation 2 for p:
p = (430 - 4b) / 2
p = 215 - 2b -------- Equation 3
Substitute Equation 3 into Equation 1:
(215 - 2b) + b + 2g = 340
215 - b + 2g = 340
-b + 2g = 125 -------- Equation 4
Now we have two equations:
-b + 2g = 125 -------- Equation 4
2p + 4b = 430 -------- Equation 2
Multiply Equation 4 by 2, so the coefficients of 'b' will cancel out:
-2b + 4g = 250 -------- Equation 5
Add Equation 5 to Equation 2:
2p + 4b + (-2b + 4g) = 430 + 250
2p + 2b + 4g = 680
Divide the equation by 2:
p + b + 2g = 340
Notice that this equation is the same as Equation 1, which means our system has infinite solutions. This happens because the number of equations is not sufficient to determine the unique values of p, b, and g.
To check this, let's plug in some values that satisfy the equations. For example, let's choose p = 100, b = 50, and g = 25:
5 * 100 + 10 * 50 + 3 * 25 = 500 + 500 + 75 = 1075
20 * 100 + 30 * 50 + 20 * 25 = 2000 + 1500 + 500 = 4000
2 * 100 + 4 * 50 = 200 + 200 = 400
These values satisfy the equations, so there are multiple possible solutions for the prices of the three items.