the sum of two different numbers is 40 . if the smaller number is divided by 4 and the larger number by 12 the sum of the quotients is 6.find the larger number
Let x represent the smaller number, and y the larger. You have two equations in 2 unknowns. Sentence one produces
1) x + y = 40
Sentence two produces
2) (x/4) + (y/12) = 6
Now you have to work to eliminate one variable.
If you multiply the entire equation 2 by -12
-12[(x/4) + (y/12) = 6]
you obtain
-3x - y = -72 then add it to number 1)
x + y = 40
You obtain
-2x = - 32, thus dividing by -2 you get x, then sub it back into an original equation and solve for y
To solve this problem, let's assign variables to the two different numbers. Let's call the smaller number "x" and the larger number "y".
According to the problem, the sum of the two numbers is 40, so we can write the equation:
x + y = 40 (Equation 1)
Next, we are given that if the smaller number is divided by 4 and the larger number is divided by 12, the sum of the quotients is 6. In other words, we have:
(x/4) + (y/12) = 6 (Equation 2)
To find the larger number, we need to solve these two equations simultaneously. Here's how:
First, let's multiply Equation 2 by 12 to get rid of the fractions:
12 * [(x/4) + (y/12)] = 12 * 6
3x + y = 72 (Equation 3)
Now we have two equations:
x + y = 40 (Equation 1)
3x + y = 72 (Equation 3)
To solve this system of equations, we'll use the method of elimination. We'll subtract Equation 1 from Equation 3:
(3x + y) - (x + y) = 72 - 40
2x = 32
Divide both sides by 2:
x = 16
Now that we've found the value of x, we can substitute it back into Equation 1 to find y:
16 + y = 40
y = 40 - 16
y = 24
Therefore, the larger number is 24.