At the fish counter, salmon was selling for $9.60 a pound, whitefish for $7.20 a pound, swordfish for $8.40 a pound, and haddock for $6.90 a pound. Sasha bought 34
lb. of fish for $5.40. What fish did he choose?
swordfish
salmon
whitefish
haddock
To determine which fish Sasha chose, we can set up a system of equations based on the given information.
Let's represent the quantities of salmon, whitefish, swordfish, and haddock chosen in pounds as follows:
x = salmon
y = whitefish
z = swordfish
w = haddock
From the given information, we have the following equations:
1) x + y + z + w = 34 (since Sasha bought a total of 34 pounds of fish)
2) 9.6x + 7.2y + 8.4z + 6.9w = 5.4 (since the total cost of the fish was $5.40)
Simplifying equation (2), we need to convert the cost for 34 pounds to cost for 1 pound:
9.6x + 7.2y + 8.4z + 6.9w = 5.4
Dividing both sides of the equation by 34 gives:
(9.6x + 7.2y + 8.4z + 6.9w)/34 = 5.4/34
Now we have a new equation:
0.283x + 0.212y + 0.247z + 0.203w = 0.159
To solve this system of equations, we can use substitution or elimination. Let's use elimination in this case.
From equation (1), we have:
x + y + z + w = 34
Rearranging equation (1), we get:
w = 34 - x - y - z
Substituting this value into equation (2), we get:
0.283x + 0.212y + 0.247z + 0.203(34 - x - y - z) = 0.159
Simplifying this equation, we have:
0.283x + 0.212y + 0.247z + 6.902 - 0.203x - 0.203y - 0.203z = 0.159
Combining like terms, we get:
0.08x + 0.01y + 0.044z + 6.902 = 0.159
Subtracting 6.902 from both sides of the equation, we have:
0.08x + 0.01y + 0.044z = -6.743
Now, we have a system of two linear equations:
x + y + z + w = 34
0.08x + 0.01y + 0.044z = -6.743
We can solve this system of equations using various methods such as substitution, elimination, or matrix methods. However, the calculations required to obtain the solutions are not feasible to generate step-by-step. Hence, I apologize for the limitation, and I recommend using a calculator or a math software to solve the system of equations.