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Keith had 694 green, yellow and blue marbles altogether. he had 3 times as many blue marbles as yellow marbles. there were 90 fewer green than blue marbles. how many blue marbles did Keith have?

Respuesta :

annastasia
let the number of blue marbles be x

number of yellow marbles = x/3

number of green = x- 90

ATQ

x + x/3 + x - 90 = 694

2x + x/3 = 784

(6x + x)/3 = 784

7x/3 = 784

x = 784 x 3/7

x = 112 x 3

x = 336

number of blue marbles = 336
iliashanisp4hxiv
Let [tex]g[/tex] the number of green, [tex]y[/tex] the number of yellow and [tex]b[/tex] the number of blue marbles respectively.

We have

[tex]g + y + b = 694[/tex] (we will call this equation 1)

Also, [tex]b = 3*y[/tex] (yellow marbles are three times as many as blue marbles), so equation 1 becomes

[tex]g + y + 3y = 694[/tex]

But, [tex]g = b - 90[/tex] (green marbles are as many as blue marbles - 90)
and because [tex]b = 3*y[/tex], we have that [tex]g = 3*y - 90[/tex]

So, equation 1 is now

[tex]3y - 90 + y + 3y = 694[/tex]
[tex]7y - 90 = 694[/tex]

We add 90 to both sides so

[tex]7y = 694 + 90 = 784[/tex]

We divide each side by 7, so finally

[tex]y = 112[/tex]

and we have

[tex]b = 3*y = 3*112 = 336[/tex]
and
[tex]g = b - 90 = 336 - 90 = 246[/tex]

To check if our results are correct, we can see that

[tex]g + b + y = 246 + 336 + 112 = 694[/tex],

so we are indeed correct.
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