Two consecutive integers are such that 7 times the biggest and then added to the smallest equals.
Answers
Please remove hash
The numbers are 22 and #23#
Explanation:
Alright, to solve a problem like this, we need to read and define as we go. Let me explain.
So we know that there are two consecutive integers. They can be #x# and #x+1#. Since their consecutive, one has to be #1# number higher (or lower) than the other.
Ok, so first we need "seven times the larger"
#7(x+1)#
Next, we need to "minus three times the smaller"
#7(x+1)-3x#
Is equal to "#95#"
#7(x+1)-3x=95#
Alright! There's the equation, now we just need to solve for #x#! First we are going to get everything on one side and distribute the #7#.
#=7x+7-3x-95#
#=4x-88#
Pull out a #4#
#=4(x-22)#
Now that we have two terms, we can set them both equal to #0# and solve.
#4!=0#
This can never be true, lets move to the next term
#(x-22)=0#
#x=22#
That's it! So your two consecutive numbers are #22# and #23#!
If you want to check this, just put #22# in place of the #x# and #23# in place of the #(x+1)# in the equation we made above!