PLease help- find 2 intergers that are consectutive positive?

Question

Two intergers are consecutive positive. If the smallest interger multiplied by 3 times the larger interger equals 36, find the 2 intergers.

Is this even possible?

Answer 1

Yes, it's possible.

let the first number be x.
the next number is therefore x+1

so, from your question, we have:

x*3*(x+1)=36
x^2+x=12
x^2+x-12=0
(x+4)(x-3)=0

so the possible x's are: -4, and 3

since it has to be a positive number,

the two numbers are 3 and 4.

The key to these kind of questions is to put them in terms of x. the solution should then be fairly obvious.

Answer 2

Let's see. Two consecutive integers would be set up as x and x + 1. The second sentence of your problem gives you
x[3(x + 1)] = 36
x[3x + 3 ] = 36
3x^2 + 3x - 36 =0 You need the 0 because it's a quadratic equation.
Now divide out the common factor of 3.

x^2 + x - 12 = 0 Now factor into two binomials.
(x-3)(x+4) = 0 Set each factor to 0 and solve.
x - 3 = 0
x = 3 The other number will be 4.

x + 4 = 0
x = -4 You can reject this answer because it's not positive.

Answer 3

3 and 4

Answer 4

Call the first integer x. The second consecutive integer is x+1.
The next sentence can be written as (x)3(x+1)=36. Expand.
(x)(3x+3)=36
3x^2+3x=36
3x^2+3x-36=0 (This makes the equation a polynomial)
Simplify by dividing by 3.
x^2+x-12=0
Find the factors.
(x+4)(x-3)=0
x=(3, -4)
Since you answer has to be postive, the answer is 3.

Just to make sure.
3 and the next consectutive positive integer 4.
3 x (3x4) = 36
3 x 12 = 36

The two consectutive positive integers are 3 and 4.

Answer 5

enable: n = ist even integer n + 2 = 2d even integer Equation: n^2 + (n + 2)^2 = a hundred n^2 + n^2 + 4n + 4 = a hundred 2n^2 + 4n + 4 = a hundred 2n^2 + 4n - ninety six = 0 2(n^2 + 2n - 40 8) = 0 n^2 + 2n - 40 8 = 0 (n + 8)(n - 6) = 0 n + 8 = 0.....or.....n - 6 = 0 n = -8 .....or..... n = 6 pick n = 6 Reject n = -8 as a result: n = 6 .......First Even Integer n + 2 = 8 ...2d Even Integer verify: n^2 + (n + 2)^2 = a hundred 6^2 + (6 + 2)^2 ?=? a hundred 36 + sixty 4 ?=? a hundred a hundred = a hundred verify as a result: the two integers are 6 and eight.

Answer 6

well let's call the numbers n and n+1.
Then the problem is n*3*(n+1) = 36
or 3n^2 + 3n = 36
or n^2 +n = 12
I'll leave it to you to figure out what n is.

Answer 7

Let x= the smaller integer
x+1=the next consecutive integer

equation should work out to be...........

x(3(x+1) ) =36

Bracket button isn't typing, so it reads x bracket 3times x+1 bracket.

distribute and solve for x