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I'm working through brilliant.org's 'using variables' section in their basic math course. I'm trying to understand the following and not sure what I am doing wrong, or maybe I'm understanding it correctly?

Here's the statement 5 (1-x) + 6x = 5-5x + 6x = 5 + x

If x=5, then in my calculations I do the following:

For the first part before the = sign: 1-5=-4 and 6x = 30, therefore 5x-4 = -20 + 30 = 10

For the 2nd part 5x = 25 so 5-25 + -20 + 30 = 10

For the 3rd part 5+5 = 10

Now I actually write it out, I see they all equal 10, is this actually correct?

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    $\begingroup$ Yes: it uses properties of multiplication and addition to show $5(1-x)+6x=5+x$ $\endgroup$
    – Henry
    Aug 5, 2017 at 23:48
  • $\begingroup$ @Henry Maybe I just got confused cause it was dealing with minus values. Thank you. I will beat math dyslexia ;) $\endgroup$ Aug 5, 2017 at 23:49

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well this is a simple equation and you can solve it by isolating the x, so yes, this is correct!

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  • $\begingroup$ Simple, unless one is dyslexic, in which case the numbers have wings. It truly makes things very difficult. Just imagine if your equation moves around like a kaleidoscope of little butterflies! ;) $\endgroup$ Aug 6, 2017 at 0:37
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    $\begingroup$ well i hope you will get more familiar with this, can you pleas mark the answer as correct? $\endgroup$ Aug 6, 2017 at 0:38
  • $\begingroup$ Well, thank you. I'm trying to ascertain whether or not it is 'correct' :) Or whether it added anything. I mean, it's certainly not wrong. I think what is challenging is that the first part will always end up with a negative number, (that threw me off). That is, after one has told the numbers to 'SIT!' like a dog, and stay where they were put. $\endgroup$ Aug 6, 2017 at 0:43
  • $\begingroup$ i do not understand what do you mean by "the first part will always end up with a negative number" $\endgroup$ Aug 6, 2017 at 0:47
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    $\begingroup$ when you try to calculate 5 (1-x) + 6x you obtain 5*(-4) + 6x so -20 + 6 * 5 so -20 + 30 so in the end it will result positive $\endgroup$ Aug 6, 2017 at 0:53

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