What is the value of x?

Hi Vips.
X=0 is just the one value of X.
The other value of the equation has to satisfy the equation \(x-5-35y=0\) for which we don't know the value of y.
The two equations are:
\(x(x-5-35y)=0\)

Hi Marcab,
the value of x=0 and y=109/5 (from statement 1 or 2)
these values satisfy both equations and are only values to do so.

I'm not sure, how did you formulate:
\(x(x-5-35y)=0\)

Best thing to do (while solving such linear equations) is to eliminate 1 term, as i did in the solution.

First given eqn: x + 35y = 109*7........... (i) x 5
we get: 175y + 5x = 763*5.......(iii)

From the 2nd given equation: 175y + 763x = 763*5....(ii);

we say (ii) minus (iii), we get 763x – 5x = (763*5) – (763*5) = 0
x = 0;
y = 109*7 / 35 = 21.8;.....(0, 21.8)
How come E?

OFFICIAL EXPLANATION
Solution: E
If we have two variables (x and y), we can generally solve for both of them if we have two independent linear equations,
i.e. two linear equations that are not multiples of each other. In a data sufficiency question like this, be skeptical as soon
as you notice that 763 (in statement (2)) just happens to be 109 (in statement (1)) times 7. Manipulate statement (2)
until you get x and y on the same side: 5x + 175y = 3815. A little mental math should show that statement (2) is just 35
times statement (1), so the two equations are NOT independent and we cannot solve for x or y; (E).

The two given equations are:
1)\(x+35y=763\)
2)\(175y=763(5-x)\) which can be written as \(175y/(5-x)=763\)

since the values on Left Hand Side in both equations are equal to 763, therefore:
\(x+35y=175y/(5-x)\)
or
\((5-x)(x+35y)=175y\)

On expanding this above equation, we'll get:
\(5x+175y-x^2+35xy=175y\)

\(5x+175y-x^2+35xy=175y\)

Now we get:

\(x^2-5x-35xy=0\)

Taking x common,
we get:

\(x(x-5-35y)=0\)
One value of x=0 and the other should satisfy the equation \(x-5-35y=0\) and for this we need to know the value of Y.

When one considers the value of x=0, then he is neglecting the other value that satisfies the equation \(x-5-35y=0\).
Hence E.
Hope that helps

The other and under 1 minute solution is:

\((1) x/7 + 5y = 109\)
or
\(x+35y=763\)

Multiply the above equation by 5:

\(5{x+35y=763}\)

That gives:

\(5x+175y=763*5\)------>Equation 1


\((2) 175y = 763(5 - x)\)

This statement can be rewritten as:

\(5x+175y=763*5\)-------> Equation 2

Equation 1 and 2 are the same, so how can they answer the question guys.

From both the solutions, the clear algebra and logic clearly tell us that the answer is E.

(1) x + 35y =109*7...(i) => 35y = 109*7 - x
(2) 35*5y = 109*7(5-x) => 35y = 109*7 – [(109*7)/5]x

combining both eqnts: 109*7 – x = 109*7 - (109*7/5)x
x - (109*7/5)*x = 0 => x = 0.
from 35y = 109*7 - x, we get y = 109*7/35 = 109/5...same as before
No other value satisfies the equation.

Implying that the OA SHOULD HAVE BEEN C (E not correct): Agree with Vips;

35y= -x + 763
y = -1/35x + 109/5: slope= -1/35..............(i)

175y = 763*5 - 763x
y = - (763/175)x + 763*5/175
y = -109/35x + 109/5: slope = -109/35........(ii)

2 LINEAR eqns with 2 different slopes can only meet at a point.

I agree with Marcab and Bull....

equ 1: \(x + 35y = 763\)
equ 2: \(175y = 763 ( 5-x)\)

So
\(175y = (x-35y) (5-x)\)

As others said single equation with 2 variables, so we cannot deduce further to solution...

moreover GMAT will not make our life tat easy by giving two direct equations (you know about that very well VIPS better than us )

Hii BULL.
You and Vips are equally right in your approach but when you are considering one value of X, you are letting the other one to slip.
The solution that you provided gives only the value of X which is other than zero, i.e. second value.

Moreover 2x+2y=10 is same as 4x+4y=20.
I hope you are getting my point.
Both the statements are the same, but written in different manner and one cannot deduce a particular value of x or y.

Hi Marcab,
am sure u can see that the 2 eqns are not the same.

Bull,
The two equations ARE same.

The second equation is

\(5x +175y=763*5\)

If you divide this by 35,
then you will get the same equation as statement 1.

Since it is an equation, we can divide by 35 on both sides comprehensively.
Just because Statement 1 has been multiplied by 35 to yield statement 2 doesnt makes it a different equation.

The two equations are not the same:

\(\frac{x}{7} + 5y = 109\) --> \(y=\frac{109}{5}-\frac{x}{35}\);

\(175y = 763(5 - x)\) --> \(y=\frac{109}{5}-\frac{109x}{25}\).

Solution: \(x=0\) and \(y=\frac{109}{5}\).

Answer: C.

Hii Bunuel, found my mistake.
But what about this solution

Why do you solve two simple linear equations this way?

That came intuitively apart from that silly and disastrous mistake in second solution, I feel comfortable with this solution because it yields two values of x.
I agree to your's, Bull's and Vips's solution but whats wrong in this method.

Try to go on and solve further.

How can I solve x-35y=5.
There can be uncountable number of values.
When:
y=1, x=40
y=2, x=75

There can be huge number of values.
Bunuel are you kidding, How can I solve these?

Try to put the two values in equation 1 in THE question, you will see your error.