Question

It is required to divide the number 48 into two such parts, that the one part may be three times as much above twenty [3(A-20)], as the other wants 20[B+20]. Find both parts.

Answer 1

To find the two parts that divide the number 48, set up the equation 3(A-20) = B+20. Simplify the equation and solve the system to find A = 32 and B = 16.

Step-by-step explanation:

To find the two parts that divide the number 48, we can set up the following equation:

3(A-20) = B+20

We can solve this equation by distributing the 3 and simplifying:

3A - 60 = B + 20

Next, we can bring like terms together:

3A - B = 80

Now, we have a system of equations. We can also use the fact that the two parts should add up to 48:

A + B = 48

We can solve this system of equations to find the values of A and B. Let's subtract the second equation from the first:

(3A - B) - (A + B) = 80 - 48

2A - 2B = 32

Dividing by 2, we get:

A - B = 16

Adding this equation to the second equation, we can eliminate B:

(A + B) + (A - B) = 48 + 16

2A = 64

Dividing by 2, we find:

A = 32

Substituting this value back into A + B = 48, we can find B:

32 + B = 48

B = 16

Therefore, the two parts that divide the number 48 are 32 and 16.