A researcher performs an experiment to test a hypothesis that involves the nutrients niacin and retinol. She feeds one group of laboratory rats a daily diet of precisely 34.53 units of niacin and 25,430 units of retinol. She uses two types of commercial pellet foods. Food A contains 0.14 unit of niacin and 130 units of retinol per gram. Food B contains 0.29 unit of niacin and 90 units of retinol per gram. How many grams of each food does she feed this group of rats each day?
food a____g
food b____g
To solve this problem, we can set up a system of equations based on the given information.
Let's assume the researcher feeds x grams of Food A and y grams of Food B each day.
According to the information given, the total amount of niacin consumed per day should be 34.53 units. Considering that Food A contains 0.14 units of niacin per gram and Food B contains 0.29 units of niacin per gram, we can write the equation:
0.14x + 0.29y = 34.53 (Equation 1)
Similarly, the total amount of retinol consumed per day should be 25,430 units. Since Food A contains 130 units of retinol per gram and Food B contains 90 units of retinol per gram, we can write the equation:
130x + 90y = 25,430 (Equation 2)
We now have a system of two equations that we can solve to find the values of x and y, representing the number of grams of Food A and Food B, respectively.
There are several methods to solve a system of equations, such as substitution or elimination. Here, we'll use the elimination method:
Multiply Equation 1 by 100 to eliminate decimals:
14x + 29y = 3453 (Equation 3)
Now, multiply Equation 2 by 10:
1300x + 900y = 25430 (Equation 4)
Next, multiply Equation 3 by 30 to get:
420x + 870y = 103590 (Equation 5)
Subtract Equation 4 from Equation 5:
420x + 870y - 1300x - 900y = 103590 - 25430
-880x - 30y = 78160
Divide through by -10:
88x + 3y = -7816 (Equation 6)
We now have a new equation (Equation 6) that we can use to eliminate one of the variables.
Multiply Equation 1 by 6:
0.84x + 1.74y = 207.18 (Equation 7)
Multiply Equation 6 by 29:
2544x + 87y = -226444 (Equation 8)
Multiply Equation 7 by 87:
73.08x + 151.38y = 18011.06 (Equation 9)
Subtract Equation 8 from Equation 9:
73.08x + 151.38y - 2544x - 87y = 18011.06 - (-226444)
-2470.92x + 64.38y = 244455.06
Divide through by -127.34:
19.41x - 0.51y = -1920.66 (Equation 10)
Now we have a new equation (Equation 10) that can be used to eliminate the other variable.
Multiply Equation 6 by 0.51:
44.88x + 1.53y = -3984.16 (Equation 11)
Multiply Equation 10 by 1.74:
33.79x - 0.89y = -3348.42 (Equation 12)
Subtract Equation 11 from Equation 12:
33.79x - 0.89y - 44.88x - 1.53y = -3348.42 - (-3984.16)
-11.09x + 0.64y = 635.74
Divide through by 0.47:
-23.60x + 1.36y = 1350.76 (Equation 13)
Now we have one final equation (Equation 13) with only one variable, x.
Multiply Equation 10 by 1.36:
26.46x - 0.69y = -2611.42 (Equation 14)
Subtract Equation 14 from Equation 13:
26.46x - 0.69y - (-23.60x + 1.36y) = -2611.42 - 1350.76
50.06x - 2.05y = 1260.66
Divide through by 50.06:
x - 0.04y = 25.19
Now, we have x in terms of y:
x = 0.04y + 25.19
To find integer values for x and y, we can substitute different values for y and solve for x. The values we substitute for y should be multiples of 4 since the coefficient of y is 0.04.
Let's start by trying y = 0. When y = 0, x = 25.19. However, we want both x and y to be positive, and using y = 0 would mean the rats are not eating any Food B.
Let's try y = 4. When y = 4, x = 0.04(4) + 25.19 = 25.35.
Since both x and y should be positive, we have a solution:
food a: 25.35 grams
food b: 4 grams
This is just Algebra I.
If there are
a grams of A
b grams of B
.14a + .29b = 34.53
130a + 90b = 25430
Now just find a and b.