3 years ago

If y varies directly as the square of x and y =12 when x=2 find the value of x when y = 108

Mathematics

1 answer:

telo118 [61]3 years ago

5 0

y = 3(x^2)

108 = 3(x^2)

36 = x^2

x = 6

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Solve the system: 2x+6y=36 x+4y=20

Assoli18 [71]

Let's solve this problem step-by-step

Let's set:

2x + 6y = 36 -- equation 1

x + 4y = 20 -- equation 2

(equation 2) * 2

2x + 8y = 40 -- equation 3

(equation 3) - (equation 1)

2y = 4

y = 2 -- equation 4

Plug (equation 4)'s value of y into (equation 2)

x + 4(2) = 20

x = 20 - 8

x = 12

Thus x = 12 and y = 2

Let's check, by substituting these values

$2x + 6y = 36\\2(12)+6(2) = 36\\24+12=36\\36=36\\\\x+4y=20\\12+4(2)=20\\12+8=20\\20=20$

Answer: x = 12 and y = 2

Hope that helps!

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