SOLUTION: one number is nine less than a second number five times the first is 9 more than 3 times the second

Hi, there--

The Problem:
One number is nine less than a second number. Five times the first is 9 more than 3 times the second.
Identify the two numbers.

A solution:
Let x be the first number.
Let y be the second number.

Now write two equations using the information in the problem.
{1}
One number is nine less than a second number. 
In algebra, we write this equation as
x = y - 9

Five times the first is 9 more than 3 times the second.
In algebra, the translates to
5x = 3y + 9

Now we have a system of two equations. We'll solve the system using the elimination method.
Multiply every term in the first equation by -5.

x = y - 9 --------------> -5x = -5y + 45
5x = 3y + 9 ----------->   5x = 3y  + 9

Add the two equations together. Notice that -5x + 5x = 0, so we no longer have an x-term.
0 = -2y + 54
.
Add 2y to both sides of the equation.
2y = 54

Divide both sides of the equation by 2.
y = 27

The second number y is 27.

Substitute 27 for y in the first equation.
x = y - 9
x = (27) - 9
x = 18

The first number x is 18.

Check your work by substituting this solution into the original problem.
One number is nine less than a second number. 
18 is nine less than 27. (True!)

Five times the first is 9 more than 3 times the second.
5 times 18 is 9 more than 3 times 27.
5 times 18 is 90; 3 times 27 is 81. 90 is 9 more than 81. (True!)

Everything checks out. Feel free to email me if you have questions about the solution.

Ms.Figgy
math.in.the.vortex@gmail.com