If x and y are real numbers such that x > y > 0 and they satisfy the following system of e… — GRE General Test

x

and

y

are real numbers such that

x > y > 0

and they satisfy the following system of equations:

\frac{3}{x+y} + \frac{4}{x-y} = \frac{11}{4}

\frac{5}{x+y} - \frac{2}{x-y} = \frac{1}{4}

what is the value of

x

Cevap: 3

Cevap

The value of

x

is 3.

Substituting

u = \frac{1}{x+y}

and

v = \frac{1}{x-y}

transforms the non-linear looking equations into the linear system

3u + 4v = \frac{11}{4}

and

5u - 2v = \frac{1}{4}

. Solving this system yields

u = \frac{1}{4}

and

v = \frac{1}{2}

. Consequently,

x + y = 4

and

x - y = 2

. Adding these two equations gives

2x = 6

, so

x = 3

Adım Adım Çözüm

Introduce auxiliary variables to linearize the system.

Let

u = \frac{1}{x+y}

and

v = \frac{1}{x-y}

. The system becomes

3u + 4v = \frac{11}{4}

and

5u - 2v = \frac{1}{4}

Replacing non-linear reciprocal terms with simple variables allows elimination or substitution methods for linear systems.

Solve the system of linear equations for

u

and

v

using elimination.

Multiply

5u - 2v = \frac{1}{4}

by 2 to get

10u - 4v = \frac{1}{2}

. Add this to

3u + 4v = \frac{11}{4}

13u = \frac{11}{4} + \frac{2}{4} = \frac{13}{4} \implies u = \frac{1}{4}

. Then

4v = \frac{11}{4} - 3\left(\frac{1}{4}\right) = 2 \implies v = \frac{1}{2}

Eliminating

v

yields a single equation in

u

, which provides the values of both auxiliary variables.

Convert auxiliary values back to equations in

x

and

y

Since

u = \frac{1}{x+y} = \frac{1}{4}

, we get

x + y = 4

. Since

v = \frac{1}{x-y} = \frac{1}{2}

, we get

x - y = 2

Inverting the fractions restores the original variables in a standard 2x2 linear system.

Solve for

x

by adding the two linear equations.

(x + y) + (x - y) = 4 + 2 \implies 2x = 6 \implies x = 3

Adding the equations eliminates

y

directly, isolating

x

Anahtar Kavram

Solving systems of linear equations using substitution variables for algebraic simplification

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