How do you solve this
Find two consecutive integers such that the lesser added to three times the greater gives a sum of 46

Question

How do you solve this
Find two consecutive integers such that the lesser added to three times the greater gives a sum of 46

unklerhaukus · Answer

the the smaller of the integres be $n$
the other inter is consecutive so it is $n+1$

unklerhaukus · Answer

$$n+3(n+1)=46$$

anonymous · Answer

let x be the 1st integer(lesser)
3(x+1) be the second integer

unklerhaukus · Answer

solve for $n$

anonymous · Answer

How do you get two numbers then

unklerhaukus · Answer

$n$ is the smaller of the integres
the other is $n+1$

unklerhaukus · Answer

for some reason im not getting integer values though

anonymous · Answer

Yea I don't get it

anonymous · Answer

i guess the question is wrong

anonymous · Answer

The integers have to be even

anonymous · Answer

let the two consecutive numbers be n, n + 2
the question says if we add the lesser number $$\left( n \right)$$ to three times the greater number $$3\left( n + 2 \right)$$ we get 46

so in the form of equation we get


$$\left[ \left( n \right) + 3\left( n +2 \right) \right] = 46$$

now on solving it

$$\left[ n  + 3n +6 \right] = 46$$

$$4n + 6 = 46$$

$$4n = 40$$

$$n = 10$$

the numbers are 10 and 12

anonymous · Answer

Yea that's right thanks so much

anonymous · Answer

wheres my medal :D

anonymous · Answer

Oh sorry