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# Linear Stuff I'm stuck on

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The sum of three numbers a, b, and c is 99. If we increase a by 6, decrease b by 6 and multiply c by 5, the three resulting numbers are equal. What is the value of b?

Oct 4, 2019

#1
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a + b + c  = 99

And if

a + 6  = b - 6

Then  b = a + 12

And  a + 6  = 5c   ⇒   [a + 6] / 5  = c

So

a + b + c  =  99

a  + ( a + 12)  +  [ a  + 6 ] / 5  = 99      multiply through by 5

5a + 5a  + 60 +  a + 6  =  495      simplify

11a  + 66 = 495       subtract 66 from both sides

11a  = 429     divide both sides by 11

a  =  39

b = a+ 12  =  51   Oct 4, 2019
edited by CPhill  Oct 4, 2019
edited by CPhill  Oct 4, 2019
#2
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The sum of three numbers a, b, and c is 99. If we increase a by 6, decrease b by 6 and multiply c by 5, the three resulting numbers are equal. What is the value of b?

Die Summe der drei Zahlen a, b und c ist 99. Wenn wir a um 6 erhöhen, b um 6 verringern und c mit 5 multiplizieren, sind die drei resultierenden Zahlen gleich. Was ist der Wert von b?

$$a+b+c=99$$

$$a+6=b-6\\ 5c=b-6\\ c=\frac{b-6}{5}$$

$$a+b+\frac{b-6}{5}=99\\ a-b=-12\\ 2b+ \frac{b-6}{5}=111\\ 10b+b-6=561\\ 11b=549$$
$$b=51$$

$$a+6=b-6\\ a+6=51-6$$

$$a=39$$

$$5c=b-6\\ 5c=51-6$$

$$c=9$$ !

Oct 4, 2019
edited by asinus  Oct 4, 2019