separate 50 into 2 parts such that 3 times the smaller is 10 less than twice the larger. Find both parts.
S + L = 50
3S + 10 = 2L
2S + 2L = 100
3S + 10 = 100 - 2S
5S = 90
Let's assume the smaller part is x, and the larger part is y.
According to the given conditions:
1. "3 times the smaller is 10 less than twice the larger" can be written as: 3x = 2y - 10.
2. "Separate 50 into 2 parts" can be written as: x + y = 50.
We now have a system of two equations:
1. 3x = 2y - 10
2. x + y = 50
To solve this system, we can use the substitution method.
From equation 2, we can rewrite it as x = 50 - y.
Substituting this into equation 1, we have:
3(50 - y) = 2y - 10
150 - 3y = 2y - 10
150 + 10 = 2y + 3y
160 = 5y
y = 32
Substituting the value of y back into equation 2:
x + 32 = 50
x = 50 - 32
x = 18
Therefore, the two parts are 18 and 32.
To solve this problem, we can set up an equation based on the given information.
Let's represent the two parts as x and y.
According to the problem, we have two conditions:
1. "3 times the smaller is 10 less than twice the larger."
This can be written as: 3x = 2y - 10.
2. The sum of the parts is 50.
This can be written as: x + y = 50.
Now we have a system of two equations with two variables. We can solve it using substitution or elimination method.
Let's solve it using substitution:
From equation 2, we have x = 50 - y.
Substituting this value of x into equation 1, we get:
3(50 - y) = 2y - 10.
Simplifying this equation:
150 - 3y = 2y - 10.
Moving all terms with y to one side:
150 + 10 = 2y + 3y,
160 = 5y.
Dividing both sides by 5:
y = 32.
Substituting this value of y into equation 2:
x + 32 = 50.
Subtracting 32 from both sides:
x = 18.
Therefore, the two parts are 18 and 32, respectively.