One integer is 4 more than a second integer. Three times the first integer when added to two times the second integer is less than 23. Determine the largest possible values of the two integers.
9 divided by 66
Let x = one integer and x + 4 = the second.
3x + 2(x+4) < 23
Solve for x and x+4.
I hope this helps. Thanks for asking.
To solve this problem, let's denote the first integer as x and the second integer as y.
We are given that one integer is 4 more than the other, so we can write the equation: x = y + 4.
We are also given that three times the first integer when added to two times the second integer is less than 23, so we can write the inequality: 3x + 2y < 23.
Now, let's substitute the first equation (x = y + 4) into the second inequality:
3(y + 4) + 2y < 23
Simplifying the equation:
3y + 12 + 2y < 23
5y + 12 < 23
5y < 23 - 12
5y < 11
y < 11/5
y < 2.2
Since y must be an integer, the largest possible value for y is 2.
Now, substitute y = 2 into the first equation to find the value of x:
x = y + 4
x = 2 + 4
x = 6
Therefore, the largest possible values for the two integers are x = 6 and y = 2.