Revised: URGENT: What is the answer to this problem? 4(3x-2) – 3x < 3(1+3x) - 7?

since 0 is less than four, the statement’s true so it’s infinite solutions. if it would’ve been false it would’ve been no solution

4(3x-2) – 3x < 3(1+3x) - 7 first, expand the stuff inside the parentheses using the distribution property: 12x - 8 - 3x < 3 + 9x - 7 simplify: 9x - 8 < 9x - 4 -8 < -4 as you can see, even if you manipulate the last inequality, the truth or correctness of the statement would remain. in cases like this, you may conclude that for any real number x, the statement will always be true. therefore x = set of all real numbers

Did you ask the previous question? Have you selected a “best answer”? Well, for this one.

4(3x – 2) – 3x < 3(1 + 3x) - 7 12x - 8 -3x < 3 + 9x - 7 9x - 8 < 9x - 4 therefore 0 < 4 No, absolute values are not the answer. As you know, zero is less than 4 so that is your answer.

On solving

4(3x – 2) – 3x < 3(1 + 3x) - 7 12x - 8 -3x < 3 + 9x - 7 9x - 8 < 9x - 4 0<4 Now, what we need to do is little rational thinking. We all know 0<4 is true add 9x on both side (MATHS RULE:: you are allowed to add/substract same amount on both sides of a inequality) 9x<4+9x is true which is same as 9x-8<9x+3-7 is true which is same as 12x - 8 -3x < 3 + 9x - 7 is true 4(3x - 2) - 3x < 3(1 + 3x) - 7 is true Whenever a mathematical expression in x (ie with a variable) comes out to be TRUE its called a IDENTITY. Hence the inequality is a IDENTITY. Examples: ~ 7x+6>6x+5+x

~ (a+b)^2=a^2 +b^2 +2ab

~ (a+b)(a-b)=a^2-b^2

the solution is all numbers less than 0

It means it’s faulse. Never 0<4.

see im doing it for ya..see and learn frm me…

4(3x-2) – 3x < 3(1+3x) - 7 12x-8-3x < 3+9x-7 9x-8 < -4+9x -8+4 < -9x+9x 4 < 0 yeah dude ur wrong...!!! how is it possible... tc