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

Answer 1

Let's assume Sasha bought x pounds of salmon, y pounds of whitefish, z pounds of swordfish, and w pounds of haddock.
According to the problem, we have the following information:
x + y + z + w = 34 (Equation 1) ---> Since Sasha bought a total of 34 pounds of fish.
9.60x + 7.20y + 8.40z + 6.90w = 5.40 (Equation 2) ---> Since Sasha spent a total of $5.40.
To solve this system of equations, we can use substitution or elimination methods. In this case, we will use the substitution method.
Let's solve Equation 1 for x:
x = 34 - y - z - w
Now substitute x in Equation 2:
9.60(34 - y - z - w) + 7.20y + 8.40z + 6.90w = 5.40
326.4 - 9.60y - 9.60z - 9.60w + 7.20y + 8.40z + 6.90w = 5.40
317.52 - 2.4y - 1.20z - 2.7w = 5.40
-2.4y - 1.20z - 2.7w = 5.40 - 317.52
-2.4y - 1.20z - 2.7w = -312.12 (Equation 3)
We now have a system of three equations:
x + y + z + w = 34 (Equation 1)
-2.4y - 1.20z - 2.7w = -312.12 (Equation 3)
Let's simplify Equation 3 and divide it by -2.4:
y + 0.5z + w = 130.05 (Equation 4)
Since we have 3 variables (y, z, w) and 3 equations (Equations 1, 3, and 4), we can solve the system.
From Equation 1: y = 34 - x - z - w (Equation 5)
We can plug Equation 5 into Equation 4:
34 - x - z - w + 0.5z + w = 130.05
34 - x + 0.5z = 130.05
0.5z = 130.05 - 34 + x
0.5z = 96.05 + x
z = 192.10 + 2x (Equation 6)
Looking at Equation 6, we can see that z must be an even number. However, none of the given options for fish (salmon, whitefish, swordfish, haddock) have a price that can be divided by 2 without a decimal. Therefore, it appears that there is no solution that satisfies all the conditions of the problem.