Math Hw Help: Solve 1+x_< (less than or equal to) 3-2x.

This is a 9th grade Math Homework Problem.
Question: Solve 1+x_< (less than or equal to) 3-2x.
Please help!!! Explain with detail!!! Thanks!

1+x < 3 - 2x

2x is some amount. we can add 2x to (1+x) and it will be less than 2x + (3-2x). like if i have 10 bucks and you have 20 bucks, we both win 5 bucks, you still have more than i do.

1 + x + 2x < 3

now we put the x's together x + 2x = 3x

1 + 3x < 3

okay, subtract one from both sides same idea as before. we can add/subtract the same amount to both sides of an equation and the equation is still true ( i.e. stays the same)

3x < 3 -1

3x < 2

Ok now we divide both sides by 3. be careful when dividing, because dividing by a negative number will reverse the inequality. that is, if i divide by (-3) i get -x > -2/3. but here we have 3 is a positive number, so when we divide the inequality sign stays the same.

x < 2/3

so that's our answer, and the less than or equals sign works the same way, you can just tack it on at the end.

1+x≤3-2x
-1 -1 (subtract one from both sides to get variable (x) alone)
x≤2-2x
+2x +2x (add 2x to both sides to get rid of variable on other side)
3x≤2
÷3 ÷3 (divide both sides by 3 to isolate x)
x≤2/3

k get x by its self by subtracting x on both sides so it would look like 1_<3-x than subtract 3 since x has to equal something so know 1_
-

1+x≤3-2x
3x≤2
x≤2/3

-

subtract one. you get: x≤2-2x. add 2x. 3x≤2. divide by 3. x≤2/3