Question 62944
Q1.  Write the following as a algebraic expression and then simplify.
a.  The sum of three consecutive intergers if "X" is the first consecutive interger.
1st number: x
2nd numer: x+1
3rd number: x+1+1=x+2
sum: +
(x)+(x+1)+(x+2)
{{{highlight(3x+3)}}}
:
b.  The perimeter of a square with side length 3x+1.
P=4s; s=3x+1
P=4(3x+1)
{{{highlight(P=12x+4)}}}
:
Q2.  2(7x-1) -5x> -(-7x) +4
14x-2-5x>7x+4
9x-2>7x+4
9x-7x-2>7x-7x+4
2x-2>4
2x-2+2>4+2
2x>6
2x/2>6/2
{{{highlight(x>3)}}}
:
Q3.  Solve: -2x-5< -3 and 6x<0
-2x-5+5<-3+5  and 6x/6<0/6
-2x<2 and x<0
-2x/-2>2/-2 and x<0  flip the inequality when you divide by -2
x>-1 and x<0
{{{highlight(-1<x<0)}}}
:
Q4.  |3x+1| + 9<1
|3x+1|+9-9<1-9
|3x+1|<-8
There's no solution the absolute value of anything is positive and can't be less than 0 much less less than -8.
:
Happy Calculating!!!