A box containing 5 red balls and 3 green balls weighs 43 g. When the box contains 3 red balls and 5 red balls it weighs 37g. When it contains one ball of each it weighs 19g. What is the mass of the box and of each ball.
mass of red --- x
mass of green --y
mass of box ---z
5x + 3y + z = 43
3x + 5y + z = 37
x + y + z = 19
subtract the first two:
2x - 2y = 6
x - y = 3
subtract the last from the first:
4x + 2y = 24
2x + y = 12
Add the two x , y equations:
3x = 15
x = 5
in 2x+y = 12 ---> 10+y = 12
y = 2
in x+y+z = 19
5+2+z=19
z = 12
each red has mass of 5
each green has mass of 2
and the box has mass of 12
cool thanks
Thanks! I was struggling a lot with this question and I really appreciate the help!
To solve this problem, we need to assign variables to the mass of the box and the mass of each ball.
Let's assume:
- The mass of the box is represented by the variable B (in grams).
- The mass of each red ball is represented by the variable R (in grams).
- The mass of each green ball is represented by the variable G (in grams).
Now, let's use the given information to set up equations:
Equation 1: 5R + 3G + B = 43 (when the box contains 5 red balls and 3 green balls)
Equation 2: 3R + 5G + B = 37 (when the box contains 3 red balls and 5 green balls)
Equation 3: R + G + B = 19 (when the box contains one ball of each color)
To solve these equations, we will use a method called substitution. We will solve Equation 3 for B, then substitute that value in Equations 1 and 2.
From Equation 3, by rearranging terms, we get:
B = 19 - R - G
Now, let's substitute B in Equation 1:
5R + 3G + (19 - R - G) = 43
Simplifying this equation, we get:
4R + 2G = 24
Next, let's substitute B in Equation 2:
3R + 5G + (19 - R - G) = 37
Simplifying this equation, we get:
2R + 4G = 18
Now, we have a system of equations:
4R + 2G = 24
2R + 4G = 18
To solve this system, we can use the method of elimination.
Multiply the first equation by 2 and the second equation by -2 to eliminate G:
8R + 4G = 48
-4R - 8G = -36
Adding these equations, we get:
4R = 12
Dividing both sides by 4, we find that R = 3 grams.
Substituting this value of R into any one of the original equations (e.g., Equation 3), we find:
3 + G + B = 19
G + B = 16
Since we have two variables (G and B) and only one equation, we need more information to solve for their specific values.
Therefore, we can conclude that the mass of each red ball is 3 grams, but we do not have enough information to determine the mass of each green ball or the mass of the box.