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What is the value of x ?

(1) 6x + 4y = 30
(2) x = -(2/3)y + 5

Source: Veritas Prep, Book 08
Chapter: 4
Topic: Data Sufficiency
Question: 11
Question: Page 84
Solution: Page 86

My Solution:
Statement 1: x = (30-4y)/6
Different values of y give different values of x. Hence Not Sufficient. Answer Choice A and D are eliminated.

Statement 2: x = -(2/3)y+5
Different values of y give different values of x. Hence Not Sufficient. Answer Choice B is eliminated.

Answer Choice C and E remain.

Statement 1 and 2 together.
Together 1 and 2 are sufficient as the value of 4 from statement 1 can be substituted in statement 2. This will provide the value of x. Hence select the answer choice C. However the correct Answer is E.

Can someone please explain why "E" is correct and Not Answer Choice "C" ?

You could simply check your theory by actually solving the way you propose and you would see why E is the answer and not C.

What is the value of x ?

(1) 6x + 4y = 30. On equation two unknowns. Not sufficient.

(2) x = -(2/3)y + 5. On equation two unknowns. Not sufficient.
Notice though that x = -(2/3)y + 5 is the same as 3x = -2y + 15 --> 3x + 2y = 15 --> 6x + 4y = 30. The same equation we had in (1)

(1)+(2) We still have only one equation with two unknowns. Not sufficient.

Answer: E.

Hope it's clear.

P.S. Please read carefully and follow: new-to-the-math-forum-please-read-this-first-140445.html Pay attention to the rule #3. Also, please hide OA under the spoiler. Thank you.

Thank You for the explanation. I did not completely solve the problem after substituting the statement 1 in 2. I made an assumption that it will provide one value. Where just rearranging the statement 2 provided the answer. I think you meant the post provided at rules-for-posting-please-read-this-before-posting-133935.html and not new-to-the-math-forum-please-read-this-first-140445.html.

Is there a post on how to use the M, Fraction, Square Root, Align etc ?

Yes, I posted the wrong link.

As for your other question, check here: rules-for-posting-please-read-this-before-posting-133935.html#p1096628

Hope it helps.
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