If 5 times the smaller of two numbers is subtracted from twice the larger the result is 16. If the larger is increased by 3 times the smaller, the result is 63. Find the numbers.

let

a = smaller
b = larger

2b - 5a = 16
b + 3a = 63

2b - 5a = 16
2b + 6a = 126

11a = 110
a = 10
b = 33

If THREE times the smaller of two numbers is subtracted from twice the larger, the result is -1. If the larger is increased by two times the smaller, the result is 24. Find the numbers.

Let

sm=a*3=3a
lg=b*2=2b
2b-3a=-1-----eqn 1
b+2a=24----eqn2, then
a=7
b=10.

To solve this problem, let's assign variables to the two numbers. Let's say the smaller number is "x" and the larger number is "y".

According to the problem, "5 times the smaller of two numbers is subtracted from twice the larger, the result is 16." This can be written as:

2y - 5x = 16 ... Equation 1

The problem also states that "the larger is increased by 3 times the smaller, the result is 63." This can be written as:

y + 3x = 63 ... Equation 2

Now we have a system of two equations (Equation 1 and Equation 2) and two variables (x and y). We can solve this system using the method of substitution or elimination.

Let's solve this system of equations using the elimination method:

Multiply Equation 2 by 2 to eliminate the terms with "y":

2(y + 3x) = 2(63)
2y + 6x = 126 ... Equation 3

Now we have:

2y - 5x = 16 ... Equation 1
2y + 6x = 126 ... Equation 3

We can add these two equations together to eliminate the term "y":

(2y - 5x) + (2y + 6x) = 16 + 126
4y + x = 142 ... Equation 4

Now we have a new equation:

4y + x = 142 ... Equation 4

To eliminate the variable "x", we can subtract Equation 4 from Equation 3:

(2y + 6x) - (4y + x) = 126 - 142
-2y + 5x = -16 ... Equation 5

Now we have two equations:

4y + x = 142 ... Equation 4
-2y + 5x = -16 ... Equation 5

Now we can use these equations to solve for the values of x and y. Let's multiply Equation 4 by 2 and Equation 5 by 4 to simplify the coefficients of y:

8y + 2x = 284 ... Equation 6
-8y + 20x = -64 ... Equation 7

Now we can add Equation 6 and Equation 7 together to eliminate the variable "y":

(8y + 2x) + (-8y + 20x) = 284 - 64
22x = 220
x = 10

Now that we have the value of x, we can substitute it back into one of the original equations to solve for y. Let's use Equation 2:

y + 3x = 63
y + 3(10) = 63
y + 30 = 63
y = 33

Therefore, the two numbers are x = 10 and y = 33.