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If 3x + 2y + 5z = 10 and 6x + 4y = 12, what is the value of z? A. [m]

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Bunuel
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If 3x + 2y + 5z = 10 and 6x + 4y = 12, what is the value of z? A. [m] [#permalink] New post   28 Oct 2021, 06:15
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If \(3x + 2y + 5z = 10\) and \(6x + 4y = 12\), what is the value of z?

A. \(\frac{4}{5}\)

B. \(\frac{7}{5}\)

C. \(2\)

D. \(\frac{12}{5}\)

E. It cannot be determined from the information given.
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Re: If 3x + 2y + 5z = 10 and 6x + 4y = 12, what is the value of z? A. [m] [#permalink] New post   28 Oct 2021, 06:22
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3x+2y+5z=10
=> 3x + 2y = 10 - 5z

6x+4y=12
=> 3x + 2y = 6

Hence 6 = 10 - 5z
=> z = 4/5

Option A
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Re: If 3x + 2y + 5z = 10 and 6x + 4y = 12, what is the value of z? A. [m] [#permalink] New post   01 Nov 2021, 11:55
Divide by 2 on both sides of the equation: 6x + 4y = 12 ——-> 3x + 2y = 6

Plug into first equation:

6 + 5z = 10

5z = 4

z = 4/5

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If 3x + 2y + 5z = 10 and 6x + 4y = 12, what is the value of z? A. [m] [#permalink] New post   04 Nov 2021, 22:40
I tried to do this question the chetan2u way: That is find values instead of algebraically solving equations.

Two equations are given: 1) 6x + 4y = 12 and 2) 3x + 2y + 5z = 10.

For 1) you can put x = 1 & y = 6/4 and the equations will hold.
Put these values of x = 1 & y = 3/2 into eq (2),
You get, 3 + 3 + 5z = 10 => z = \(\frac{4}{5}\) OPTION A
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If 3x + 2y + 5z = 10 and 6x + 4y = 12, what is the value of z? A. [m] [#permalink]
04 Nov 2021, 22:40
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