Alex is using a scale that is known to have a constant error. a can of soup and a can of tuna are placed on this scale, and it reads 24 ounces. now four identical cans of soup and three identical cans of tuna are placed on an accurate scale, and a weight of 80 ounces is recorded. if two cans of tuna weigh 18 ounces on the bad scale, then what is the amount of error in the scale and what is the correct weight of each type of can?

Let the soup can be represented by = s

Let the tuna can be represented by = t

Let the error of the scale in grams be represented by = e

4 cans of soup and 3 cans of tuna are placed on a correct scale and reads 80 ounces.

A can of soup and a can of tuna are placed on this scale, and it reads 24 ounces.

Two cans of tuna weigh 18 ounces on the bad scale.

Now equations becomes:

  ....(1)

   .... (2)

    ..... (3)

From equation 3 we get

Putting this in equation 2

=   ..... (4)  

Now we will solve equations 1 and 4.

Multiplying equation (4) by '4' we get,

Now subtracting this from equation 1, we get

Now 4s+3t=80 ,

s = 14

And s+t+e = 24

14+8+e = 24

e = 2

Hence, a can of tuna weighs 8 ounces, a can of soup weighs 14 ounces and the amount of error is 2 ounces.

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer

the answer to the expression is 83