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2 years ago

14

There are y cupcakes in the class. It is two more than 4 times the number of donuts (X). How many donuts are there in the class?

1 answer:

Snezhnost [94]2 years ago

6 0

Answer:

(c) x = (y -2)/4

Step-by-step explanation:

The given relationship between x and y can be solved for x.

__

<h3>given relation</h3>

There are 2 more cupcakes (y) than 4 times the number of donuts (x):

y = 2 +4x

<h3>solve for x</h3>

y -2 = 4x . . . . . subtract 2

x = (y -2)/4 . . . . divide by the coefficient of x . . . . matches 3rd choice

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$\bold{\huge{\orange{\underline{ Solution }}}}$

$\bold{\underline{ Given \: Rules}}$

<u>The </u><u>sum</u><u> </u><u>of </u><u>the </u><u>number </u><u>in </u><u>each </u><u>of </u><u>the </u><u>four </u><u>rows </u><u>is </u><u>the </u><u>same </u>
<u>The </u><u>sum </u><u>of </u><u>the </u><u>numbers </u><u>in </u><u>each </u><u>of </u><u>the </u><u>three </u><u>columns </u><u>is </u><u>the </u><u>same</u>
<u>The </u><u>sum </u><u>of </u><u>any </u><u>row </u><u>does </u><u>not </u><u>equal </u><u>the </u><u>sum </u><u>of </u><u>any </u><u>column </u>

$\bold{\underline{ Let's \: Begin}}$

<u>According </u><u>to </u><u>the </u><u>Second</u><u> </u><u>rule </u><u>:</u><u>-</u>

$\sf{ 75+b+83=76+80+d=a+81+85+78+c+e }$

$\sf{ 158 + b = 156 + d = 166 + a = 78 + c + e ...(1)}$

<u>According </u><u>to </u><u>the </u><u>first </u><u>rule </u><u>:</u><u>-</u><u> </u>

$\sf{ 75+76+a+78 = b+80+81+c = 83+86+d+e}$

$\sf{ 229 + a = 161 + b + c = 168 + d + e ...(2)}$

<u>From </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u><u> </u><u>we </u><u>got </u><u>:</u><u>-</u>

$\sf{ 158 + b = 166 + a, 156 + d = 166 + a }$

$\sf{ b = 166 - 158 + a, d = 166 - 156 + a }$

$\sf{ b = 8 + a, d = 10 + a ...(3)}$

<u>Subsitute </u><u>(</u><u>3</u><u>)</u><u> </u><u>in </u><u>(</u><u> </u><u>2</u><u> </u><u>)</u><u> </u><u>:</u><u>-</u>

$\sf{229+a = 161+8+a+c = 168+10+a+e}$

$\sf{ 229+a = 169+a+c = 178+a+e}$

<u>We</u><u> </u><u>can </u><u>write </u><u>it </u><u>as </u><u>:</u><u>-</u>

$\sf{ 229+a = 169+a+c \:or\:229+a = 178+a+e}$

$\sf{ c = 299-169+a-a\:or\:e = 229-178+a-a}$

$\sf{ c = 60 \: and \: e = 51 }$

<u>Subsitute </u><u>the </u><u>value </u><u>of </u><u>c </u><u>and </u><u>e </u><u>in </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u><u>:</u><u>-</u>

$\sf{ 158 + b = 156 + d = 166 + a = 78 + 60 + 51 }$

$\sf{ 158 + b = 156 + d = 166 + a = 189}$

<u>Now</u><u>, </u>

$\sf{ For \: b, 158 + b = 189 }$

$\sf{ b = 189 - 158 }$

$\sf{ b = 31}$

$\sf{ For \: d , 156 + b = 189 }$

$\sf{ d = 189 - 156 }$

$\sf{ d = 33}$

$\sf{ For \: a, 166 + a = 189 }$

$\sf{ a = 189 - 166 }$

$\sf{ a = 23 }$

Hence, The value of a, b, c, d and e is 23, 31 ,60 ,33 and 51 .

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