The sum of two numbers is 48. Four times the smaller number is equal to twice

the larger number. Find the two numbers.

Ah, the math clown is here to entertain you! Let's solve this riddle, shall we?

Let's represent the two numbers as 'x' (smaller number) and 'y' (larger number).

We are given two pieces of information:
1. The sum of two numbers is 48.
2. Four times the smaller number is equal to twice the larger number.

Using these clues, we can form two equations:

Equation 1: x + y = 48 (since the sum of the numbers is 48)
Equation 2: 4x = 2y (since four times the smaller number is equal to twice the larger number)

Now, let's solve this clownish riddle. We'll start by isolating x in Equation 2:

4x = 2y
Divide both sides by 2: 2x = y (equation 3)

Now let's substitute equation 3 into equation 1 to find the value of x:

x + 2x = 48
3x = 48
Divide both sides by 3: x = 16

So, the smaller number, x, is equal to 16.

Now we can substitute the value of x back into equation 3 to find the value of y:

2x = y
2(16) = y
32 = y

Therefore, the larger number, y, is equal to 32.

In conclusion, the two numbers are 16 and 32. And hey, the math clown strikes again with a solution! Ta-da! 🎉

Let's solve this problem step by step.

Let's call the smaller number x and the larger number y.

According to the problem, the sum of the two numbers is 48:

x + y = 48 (Equation 1)

It is also given that four times the smaller number is equal to twice the larger number:

4x = 2y (Equation 2)

To solve these equations simultaneously, we can rearrange Equation 2 in terms of x:

x = (2y) / 4
x = y / 2

Now we can substitute this value of x in Equation 1:

(y / 2) + y = 48

Multiplying both sides of the equation by 2 to eliminate the fraction:

2(y / 2) + 2y = 2 * 48
y + 2y = 96
3y = 96
y = 96 / 3
y = 32

Now that we have the value of y, we can substitute it back into Equation 1 to find x:

x + 32 = 48
x = 48 - 32
x = 16

Therefore, the two numbers are x = 16 and y = 32.

To solve this problem, we need to set up a system of equations. Let's say the smaller number is x, and the larger number is y.

According to the given information, we have two equations:

1. The sum of two numbers is 48: x + y = 48
2. Four times the smaller number is equal to twice the larger number: 4x = 2y

To find the values of x and y, we can use substitution or elimination method.

Let's go with the substitution method:

From the second equation, let's solve it for y:
y = 2x/4
y = x/2

Now, substitute this value of y in the first equation:
x + y = 48
x + x/2 = 48
Multiplying both sides by 2 to eliminate the fraction:
2x + x = 96
3x = 96
x = 96/3
x = 32

Now, substitute the value of x back into one of the original equations to solve for y:
y = x/2
y = 32/2
y = 16

So, the two numbers are 32 and 16.

let the smaller number be x,

then the larger number is 48-x

"Four times the smaller number is equal to twice the larger number"

translated into math....

4x = 2(48-x)

solve

4x = 96 - 2x
6x = 96
x = 16

So the smaller is 16, the larger is 32

check: is 4(16) = 2(32) ?? yes!