Three nuts and six bolts have a mass of 72g . Four nuts and five bolts have a mass of 66f. Find the mass of a nut and that of a bolt
Let the mass of a nut be represented by "x" and the mass of a bolt be represented by "y".
From the given information, we can create the following two equations:
3x + 6y = 72 (Equation 1)
4x + 5y = 66 (Equation 2)
To solve this system of equations, we can use the method of substitution or elimination. Let's solve it using the elimination method:
Multiply Equation 1 by 4 and Equation 2 by 3 to eliminate the "x" term:
12x + 24y = 288 (Equation 3)
12x + 15y = 198 (Equation 4)
Subtract Equation 4 from Equation 3:
(12x - 12x) + (24y - 15y) = 288 - 198
9y = 90
y = 10
Substitute the value of "y" into Equation 1 or Equation 2 to find the value of "x".
Using Equation 1:
3x + 6(10) = 72
3x + 60 = 72
3x = 72 - 60
3x = 12
x = 4
Therefore, the mass of a nut is 4g and the mass of a bolt is 10g.
Let's assume the mass of a nut is represented by 'x' and the mass of a bolt is represented by 'y'.
According to the given information:
Three nuts and six bolts have a mass of 72g.
This can be expressed as:
3x + 6y = 72 (Equation 1)
Four nuts and five bolts have a mass of 66g.
This can be expressed as:
4x + 5y = 66 (Equation 2)
Now we have a system of two linear equations with two variables.
To solve this system, we can use the method of substitution or elimination.
Let's solve it using the method of elimination:
Multiply Equation 1 by 4 and Equation 2 by 3 to eliminate the variable 'x':
12x + 24y = 288 (Equation 3)
12x + 15y = 198 (Equation 4)
Subtract Equation 4 from Equation 3 to eliminate 'x':
12x - 12x + 24y - 15y = 288 - 198
9y = 90
Divide both sides of the equation by 9:
y = 10
Now substitute the value of 'y' in Equation 1 (or 2):
3x + 6y = 72
3x + 6(10) = 72
3x + 60 = 72
3x = 72 - 60
3x = 12
Divide both sides of the equation by 3:
x = 4
Therefore, the mass of a nut (x) is 4g, and the mass of a bolt (y) is 10g.
To solve this problem, let's assign variables to the unknowns:
Let's assume the mass of a nut is represented by 'n' (in grams).
Similarly, let's assume the mass of a bolt is represented by 'b' (in grams).
We are given two pieces of information:
1) Three nuts and six bolts have a total mass of 72 grams.
Thus, we can write the equation: 3n + 6b = 72.
2) Four nuts and five bolts have a total mass of 66 grams.
So, the second equation is: 4n + 5b = 66.
Now, we have a system of two equations with two unknowns. We can solve this system of equations to find the values of 'n' and 'b'.
One way to solve this system is by using the method of substitution or elimination.
Let's use the elimination method:
Multiply the first equation by 4 and the second equation by 3 to make the coefficients of 'n' equal:
12n + 24b = 288 (equation 1)
12n + 15b = 198 (equation 2)
Now, subtract equation 2 from equation 1:
(12n + 24b) - (12n + 15b) = 288 - 198
This simplifies to:
9b = 90
Divide both sides by 9:
b = 10
Now, substitute the value of 'b' back into any of the original equations (let's use equation 1):
3n + 6(10) = 72
3n + 60 = 72
3n = 72 - 60
3n = 12
Divide both sides by 3:
n = 4
Therefore, the mass of a nut is 4 grams, and the mass of a bolt is 10 grams.