Hitomi has a total of 30 red and blue bowling balls ate her bowling alley. Each red ball weighs 8 lb. Each blue ball weighs 9 lb. If the total mass of the blue balls is 49 lb. heavier than the total mass of the red balls, how many red balls does she have?

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To solve this problem, let's first set up some equations.

Let's assume that Hitomi has "x" red balls and "y" blue balls.

Given that the total number of balls is 30, we can write the equation: x + y = 30

We are also given that the total weight of the red balls is 8x, and the total weight of the blue balls is 9y.

According to the problem, the total mass of the blue balls is 49 lb. heavier than the total mass of the red balls. Mathematically, this can be written as: 9y = 8x + 49

Now, we have a system of two equations that we can solve simultaneously to find the values of x and y.

Let's solve the system of equations:

1) x + y = 30
2) 9y = 8x + 49

From the first equation, we can isolate x: x = 30 - y

Substituting this value of x into the second equation:

9y = 8(30 - y) + 49

Expanding and simplifying:

9y = 240 - 8y + 49
9y + 8y = 240 + 49
17y = 289

Dividing both sides by 17:

y = 289 / 17
y ≈ 17

So, Hitomi has approximately 17 blue balls.

Now, substituting this value of y back into the first equation:

x + 17 = 30
x = 30 - 17
x = 13

Therefore, Hitomi has 13 red balls.

In conclusion, Hitomi has 13 red balls at her bowling alley.