What is the solution (x, y) of the following system of equations?

Question

What is the solution ( x, y ) of the following system of equations?

\frac{7}{x+y} + \frac{3}{x-y} = 1  and  \frac{1}{x+y} - \frac{2}{x-y} = 5

A. ( \frac{3}{4}, \frac{1}{4} )

B. ( \frac{3}{5}, \frac{1}{4} )

C. ( \frac{1}{4}, \frac{3}{4} )

D. ( \frac{3}{5}, \frac{1}{5} )

E. ( \frac{3}{4}, \frac{1}{5} )

GMATGuruNY · Accepted Answer

ALWAYS KEEP YOUR EYE ON THE ANSWER CHOICES.

The correct answer must satisfy the following equation:
 \frac{7}{x+y} + \frac{3}{x-y} = 1 
In every answer choice,  x+y≤1 , with the result that  \frac{7}{x+y}≥7 .
Implication:
For left side of the equation to sum to 1,  \frac{3}{x-y}  must be NEGATIVE.
 \frac{3}{x-y}  will be negative only if  x<y .
The correct answer must be C:
 \frac{1}{4} < \frac{3}{4}

C

MathRevolution · Answer

=>

Assume  A = \frac{1}{(x+y)}  and  B = \frac{1}{(x-y)}.

Then we have  7A + 3B = 1  and  A – 2B = 5  or  A = 2B + 5.

Substituting the second equation into the first gives us  7(2B+5)+ 3B = 1, 17B + 35 = 1, 17B = -34 , and  B = -2.

Substituting  B = -2  into  A = 2B + 5  gives us  A = 2(-2) + 5, A = 1.

Then  x + y = \frac{1}{A} = \frac{1}{1} = 1  and  x – y = \frac{1}{B} = \frac{1}{(-2)} = \frac{-1}{2}.

Adding  x + y = 1  and  x - y = \frac{-1}{2}  gives us  x + y + x - 7 = 1 - \frac{1}{2} .

We have  2x = \frac{1}{2}  or  x = \frac{1}{4}.

Then we have  y = 1 – x = \frac{3}{4}.

Therefore, the answer is C.
Answer: C

devavrat · Answer

Hi can you pls explain this in detail. I did not understand how you reached x+y <= 1

GMATGuruNY · Answer

The answer choices represent options for (x, y).

Every option yields a sum for x and y that is less than or equal to 1:
A -->  \frac{3}{4} + \frac{1}{4} = 1 
B -->  \frac{3}{5} + \frac{1}{4} = \frac{17}{20}  
C -->  \frac{1}{4} + \frac{3}{4}  = 1
D -->  \frac{3}{5} + \frac{1}{5} = \frac{4}{5}  
E -->  \frac{3}{4} + \frac{1}{5} = \frac{19}{20}

QuantMadeEasy · Answer

1/(x+y) = a, 1/(x-y) = b
7a + 3b = a
a - 2b = 5
Solving for a and b
a = 1, b = -2

x + y = 1
x - y = 1/2

x = 1/4
y = 3/4

C is correct.

Ahmed9955 · Answer

Consider x+y=u & x-y =v
Solve for => 7/u +3/v = 1 ....Eq1
 1/v -2/u = 5. .....Eq2
 Solve for u & v
Then solve for x & y.
Correct ans C

Posted from my mobile device

nisthagupta28 · Answer

Adding x+y=1
x+y=1
 and x−y=−12
x−y=−1/2
 gives us x+y+x−7=1−12
x+y+x−7=1−1/2.

Can someone explain this step?

AdarshSambare · Answer

Simply substitute the answer choices into the question.

If a pair like (1/4, 3/4) satisfies the first equation, that’s enough.

If time permits, quickly check the second equation for confirmation.

Otherwise, mark it and move on.

luisdicampo · Answer

Method 1: Inspecting the Answer Choices (Back-Solving) 
This is often the fastest method for systems of equations with complex fractions.

Let's look at the term  x + y  in the denominators.
In options (A), (B), (C), and (E), the sum  x + y  is a nice integer.
For example, in (C):  x = \frac{1}{4}, y = \frac{3}{4} \implies x + y = 1 .

Let's test  Option (C) :  (\frac{1}{4}, \frac{3}{4}) 
1. Calculate the denominators:
  x + y = \frac{1}{4} + \frac{3}{4} = 1 
  x - y = \frac{1}{4} - \frac{3}{4} = -\frac{2}{4} = -\frac{1}{2}

2. Plug these into the second equation (it looks slightly easier to check mentally):
  \frac{1}{x+y} - \frac{2}{x-y} = 5 
  \frac{1}{1} - \frac{2}{-1/2} 
 Dividing by  -1/2  is the same as multiplying by  -2 :
  1 - (2 	imes -2) = 1 - (-4) = 1 + 4 = 5 
 This matches the equation perfectly!

3. Quickly verify with the first equation:
  \frac{7}{x+y} + \frac{3}{x-y} = 1 
  \frac{7}{1} + \frac{3}{-1/2} 
  7 + (3 	imes -2) = 7 - 6 = 1 
 Matches.

Option (C) is correct.

Method 2: Algebraic Substitution 
If you prefer solving it directly, substitute variables to clear the fractions.
Let  u = \frac{1}{x+y}  and  v = \frac{1}{x-y} .

The equations become:
1)  7u + 3v = 1 
2)  u - 2v = 5 \implies u = 5 + 2v

Substitute (2) into (1):
 7(5 + 2v) + 3v = 1 
 35 + 14v + 3v = 1 
 17v = -34 
 v = -2

Now find  u :
 u = 5 + 2(-2) = 1

Now convert back to  x  and  y :
 \frac{1}{x+y} = 1 \implies x + y = 1 
 \frac{1}{x-y} = -2 \implies x - y = -\frac{1}{2}

Add the two linear equations:
 2x = 1 - 0.5 
 2x = 0.5 
 x = 0.25 = \frac{1}{4}

Since  x + y = 1 :
 \frac{1}{4} + y = 1 \implies y = \frac{3}{4}

Answer: C

What is the solution (x, y) of the following system of equations?

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