Three times the greater of two consecutive odd integers is 15 less than two-thirds of the lesser integer. What are the integers?

Question

Need help badly

Answer 1

Let x = the lesser of two consecutive odd integers

Then x + 2 = the greater of the two (consecutive odd integers are 2 apart, for example, 5 and 7.

Three times the greater would then be represented by 3 (x + 2).

Two thirds of the lesser would be (2/3) x and 15 less than that is (2/3)x - 15.

But those values are given as equal so the equation that models the situation:

3(x + 2) = (2/3)x - 15

There are several different ways to solve from here. One way: multiply every term by 3 to clear the equation of fractions.

9(x + 2) = 2x - 45

9x + 18 = 2x - 45

7x = -63

x = -9

x+2 = -7

Those are indeed odd integers.

Three times the greater: 3(-7) = -21

15 less than 2/3 of the lesser is: 2/3 of -9 = -6 and 15 less than that - 6 - 15 = -21

Answer 2

Let x = first odd integer, the lesser integer

x+2 = second consecutive odd integer, the greater integer

3 * (x + 2) = (2/3) * x - 15

Distributing left side:

3 x + 6

Gives us the equation 3x + 6 = (2/3)x - 15

-6 - 6

=> 3x = (2/3)x - 21

-(2/3)x -(2/3)x

=> (2 1/3)x = -21

(2 1/3)x == (7/3)x

(7/3)x = -21

multiply both sides by (3/7) which is reciprocal of (7/3)

x = -21 (3/7)

x = -9, lesser integer

x+2 = -7, greater integer

Check:

3 * greater integer = -21

2/3 lesser integer = -6

2/3 lesser integer - 15 = -21