Q) one number is 9 less than three times a second number. If the first number plus twice the second number is 16, find the two numbers.
So, here's the question in mathematical form:
3x − 9 = y
x + 2y = 16
(34/7, 39/7) is my guess (solved by substitution).
Are there any options?
The textbook answer is x = 5 & y=6 or the other way around.
To solve this problem, let's introduce two variables to represent the two numbers. Let's call the first number "x" and the second number "y".
From the given information, we have two equations:
1) "One number is 9 less than three times the second number":
x = 3y - 9
2) "The first number plus twice the second number is 16":
x + 2y = 16
Now, we can use these two equations to find the values of x and y.
To eliminate the variable x, we can rearrange the first equation and substitute it into the second equation:
x = 3y - 9
x - 3y = -9
Substituting this value of x into the second equation:
(3y - 9) + 2y = 16
5y - 9 = 16
5y = 25
y = 5
Now that we have found the value of y, we can substitute it back into either equation to find the value of x. Let's use the first equation:
x = 3y - 9
x = 3(5) - 9
x = 15 - 9
x = 6
So, the two numbers are x = 6 and y = 5.