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Hello! I have a math quiz coming up on Monday and I got two questions;

1. Is this right so far?

2. Once I get to 2 fractions, what am I suppose to do? I am confused and the teacher has provided other 2 questions like this.

- Thank you for your help, it means a lot to me!

Guest Sep 9, 2017

#1**+2 **

2(x - 3) = (-1/2)(6 + 8x)

2x - 6 = -3 - 4x <-- I think that you did this step incorrectly.

6x - 6 = -3

6x = 3

x = 3/6 = 1/2

When you multiply a fraction times a whole number, consider the whole number to be a fraction with denominator 1:

(-1/2) x (6) = (-1/2) x (6/1) = (-1 x 6) / (2 x 1) = -6 / 2 = -3

geno3141 Sep 9, 2017

#2**+2 **

I am confused on how I got that wrong? Am I suppose to convert the fraction into decimals, for the 1/2 in the start? And I am really confused on the last part you sent or else I understand the other concepts. Also thank you very much, I really appreciate it.

Guest Sep 9, 2017

edited by
Guest
Sep 9, 2017

#3**+2 **

No you do not convert it to decimals.

I'm just going to simplify the right side of the second line for you.

Because that is the bit you really messed up on.

\(-\frac{1}{2}(6+8x)\\ =-\frac{1}{2}*6\;\;+\;\;-\frac{1}{2}*8x\\ =-\frac{6}{2}\;\;+\;\;-\frac{8x}{2}\\ =-\frac{6}{2}\;\;\;\;\;-\frac{8x}{2}\\ =-3\;\;\;\;\;-\frac{4x}{1}\\ =-3\;\;\;\;\;-4x\\ =-4x-3\)

Melody
Sep 9, 2017

#4**+2 **

If I am seeing this correctly...... your first two steps are:

\(2(x-3)=-\frac12(6+8x)\) This is the given equation, and you want to solve for x .

\(2x-6=-\frac6{12}+\frac8{16}x\)

You have distributed the 2 on the left side of the equation correctly, but the right side should be...

\(2(x-3)=-\frac12(6+8x) \\~\\ 2x-6=(-\frac12)(6)+(-\frac12)(8x) \\~\\ 2x-6=(-\frac12)(\frac61)+(-\frac12)(\frac{8}{1})(x) \\~\\ 2x-6=-\frac{(1)(6)}{(2)(1)}+-\frac{(1)(8)}{(2)(1)}(x) \\~\\ 2x-6=-\frac62+-\frac{8}{2}x \\~\\ 2x-6=-3+-4x \\~\\ 2x-6=-3-4x\)

.hectictar Sep 9, 2017