Problem 1 Evaluate the expression 1/2x + 14 if x = 23. I got 25.5 Problem 2 14x = 2 (5x + 14) I got 3.5 Problem 3 (x + 20) + (2x + 10) = 90 I got 20. Please tell/show me your steps. I want to make sure that I did it right.
Is the first one supposed to say "if x = 23" instead of "if x - 23" ?
Yes, sorry about that.
also, is problem 1: (1/2x) + 14 or (1/2)x + 14
They didn't put any parenthesis. I assume it would be (1/2x).
Does it look like:\[\frac{ 1 }{ 2 }x \] or \[\frac{ 1 }{ 2x }\]
The first one.
so, (1/2)x when x = 23 is (23/2) = 11.5 and (1/2)x+14 = (1/2)(23) + 14 = 25.5
Problem 1 Evaluate the expression 1/2x + 14 if x = 23. \[\frac{ 1 }{ 2 }(23)+14\] 11.5+14 = 25.5
if you understand that one, you should be able to give the other problems a try. If not, speak up :)
@Jake: I already worked on the problems I stated in the original question. I'm just confirming that I did it correctly. :)
Thank you both for question 1. :)
Problem 2 14x = 2 (5x + 14) Distribute the 2 first 14x = 10x + 28 then isolate the variables to only one side of the equation by subtacting 10x from both sides 14x = 10x + 28 -10x = -10x ----- ------ 4x = 0 + 28 OR 4x=28 Then, divide both sides by the number which is glued to the variable...in this case 4 4x = 28 --- ---- 4 4 --------------- 1x = 7 OR x= 7 Solve for x? Answer is 7
Oh okay. I forgot to distribute the 2 in problem 2. Thank you very much! I am positive about problem 3.
Since the commutitative property states that addition can be in any which order than the parenthesis can be removed. (x + 20) + (2x + 10) = 90 is now x + 20 + 2x + 10 = 90 reorganize and combine like terms (1)x + 2x + 20 + 10 = 90 = = = 3x +30 = 90 New problem looks like 3x +30 = 90 isolate the number that is glued to the variable on one side of the equation by subtracting 30 from both sides 3x +30 = 90 -30 -30 ---- ---- 3x +0 = 60 OR 3x = 60 Then, divide both sides by the number which is glued to the variable...in this case 3 3x = 60 -- --- 3 3 ------------ 1x = 20 OR x = 20 solve for x? answer is 20
Okay, thanks! :) I got the same answer as well.
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