6. Half a pepperoni pizza plus three fourths of a ham-and-pineapple pizza contains 765 Calories. One fourth
of a pepperoni pizza plus a whole ham-and-pineapple pizza contains 745 Calories. How many Calories
are in a whole pepperoni pizza? How many Calories are in a whole ham-and-pineapple pizza?
(1 point)
600 Calories, 550 Calories
480 Calories, 640 Calories
520 Calories, 680 Calories
660 Calories, 580 Calories
0.5 P + 0.75 H = 765
0.25 P + 1.0 H = 745 multiply this by 2
0.5 P + 0.75 H = 765
0.5 P + 2.0 H = 1490
_____________________ subtract
0 P - 1.25 H = - 725
H = 580
0.5 P + 2.0 (580) = 1490
P = 660
Translate the English to Math
Pepperoni pizza --- p
ham-pineapple ---- h
"Half a pepperoni pizza plus three fourths of a ham-and-pineapple pizza contains 765 Calories"
----> (1/2)p + (3/4)h = 765
2p + 3h = 3060 **
"One fourth
of a pepperoni pizza plus a whole ham-and-pineapple pizza contains 745 Calories"
---> (1/4)p + h = 745
p + 4h = 2980 or p = 2980-4h ***
sub into **
2(2980-4h) + 3h = 3060
5960 - 8h + 3h = 3060
-5h = -2900
h = 580
sub into *** to find the number of calories in the other pizza
To solve this problem, we need to set up a system of equations based on the given information. Let's denote the number of calories in a whole pepperoni pizza as P and the number of calories in a whole ham-and-pineapple pizza as H.
From the first equation, "Half a pepperoni pizza plus three fourths of a ham-and-pineapple pizza contains 765 Calories," we can write the equation as follows:
(1/2)P + (3/4)H = 765 .......... (Equation 1)
From the second equation, "One fourth of a pepperoni pizza plus a whole ham-and-pineapple pizza contains 745 Calories," we can write the equation as:
(1/4)P + H = 745 ............... (Equation 2)
To solve this system of equations, we can use an algebraic method, such as substitution or elimination. Let's solve it using the elimination method:
Multiply Equation 2 by 4 to eliminate fractions:
4*(1/4)P + 4H = 4*745
P + 4H = 2980 .............. (Equation 3)
Now we can subtract Equation 1 from Equation 3 to eliminate P:
(P + 4H) - [(1/2)P + (3/4)H] = 2980 - 765
P + 4H - (1/2)P - (3/4)H = 2215 ............... (Equation 4)
Combine like terms:
(1 - 1/2)P + (4 - 3/4)H = 2215
(1/2)P + (13/4)H = 2215 .............. (Equation 5)
Now we have a new equation (Equation 5) that relates only P and H.
From here, we can solve the system by multiplying Equation 1 by 2 to obtain a similar coefficient for P:
2[(1/2)P + (3/4)H] = 2(765)
P + (3/2)H = 1530 .............. (Equation 6)
Now we have two equations, Equation 5 and Equation 6, which can be solved using the elimination method. Subtract Equation 5 from Equation 6 to eliminate H:
(P + (3/2)H) - ((1/2)P + (13/4)H) = 1530 - 2215
P + (3/2)H - (1/2)P - (13/4)H = -685 .......... (Equation 7)
Combine like terms:
(1 - 1/2)P + (3/2 - 13/4)H = -685
(1/2)P + (2/4 - 13/4)H = -685
(1/2)P - (11/4)H = -685 ............. (Equation 8)
Now we have a new equation (Equation 8) that relates only P and H.
To solve the system of equations, we can use the elimination method again. Multiply Equation 7 by 2 to obtain a similar coefficient for P:
2[(1/2)P - (11/4)H] = 2(-685)
P - (11/2)H = -1370 ............... (Equation 9)
Now we have two equations, Equation 8 and Equation 9, with a similar coefficient for P. Add Equation 8 to Equation 9 to eliminate P:
(P - (11/2)H) + ((1/2)P - (11/4)H) = -1370 - 685
P - (11/2)H + (1/2)P - (11/4)H = -2055 ........... (Equation 10)
Combine like terms:
(1 + 1/2)P + (-11/2 - 11/4)H = -2055
(3/2)P + (-22/4)H = -2055
(3/2)P - (11/2)H = -2055 ............... (Equation 11)
Now we have a new equation (Equation 11) that relates only P and H.
We can solve this equation for either P or H. Let's solve it for P:
(3/2)P - (11/2)H = -2055
(3/2)P = (11/2)H - 2055
P = (11/3)H - 1370 ................ (Equation 12)
Now we can substitute Equation 12 into Equation 1 to solve for H:
(1/2)((11/3)H - 1370) + (3/4)H = 765
(11/6)H - 685 + (3/4)H = 765
[(11/6)H + (3/4)H] = 765 + 685
[(44 + 18)/24]H = 1450
(62/24)H = 1450
H = 1450 * (24/62)
H = 561.29
We found that the number of calories in a whole ham-and-pineapple pizza is approximately 561.29 (rounding to the nearest calorie).
To find the number of calories in a whole pepperoni pizza, we can substitute the value of H into Equation 2:
(1/4)P + 561.29 = 745
(1/4)P = 745 - 561.29
(1/4)P = 183.71
P = 183.71 * 4
P = 734.84
We found that the number of calories in a whole pepperoni pizza is approximately 734.84 (rounding to the nearest calorie).
Therefore, the answer is that a whole pepperoni pizza contains approximately 735 Calories, while a whole ham-and-pineapple pizza contains approximately 561 Calories. Thus, the closest answer choice is "660 Calories, 580 Calories."
Whew !
We agree :)