Question

5/x+y-2/x-y=-1
15/x+y+7/x-y=10 where x is not = 0,y is not = 0

Answer 1

Answer:

x=3 and y=2

Step-by-step explanation:

Given Equation:

$\dfrac{5}{x+y}-\dfrac{2}{x-y}=-1$

$\dfrac{15}{x+y}+\dfrac{7}{x-y}=10$

$\dfrac{1}{x+y}=u$

$\dfrac{1}{x-y}=v$

Reducible linear equation

$5u-2v=-1$

$15u+7v=10$

Using elimination method to solve for u and v

We will make coefficient of u same in both equation. So, we multiply first equation by 3 and we get

15u-6v=-3

15u+7v=10

Subtract (2) - (1)

7v+6v=10+3

13v=13

v=1

Substitute v into 15u+7v=10

$15u+7(1)=10[/tex[</p><p>[tex]u=\dfrac{1}{5}$

$\dfrac{1}{x+y}=\dfrac{1}{5}\Rightarrow x+y=5$

$\dfrac{1}{x-y}=1\Rightarrow x-y=1$

Add both equation and eliminate y

2x=6

x=3

Substitute x=3 into x+y=5

3+y=5

y=2

Hence, x=3 and y=2 is solution of system of equation.

Answer 2

"Given:

$\frac { 5 }{ x+y } -\quad \frac { 2 }{ x-y } = -1\\ \frac { 15 }{ x+y } +\frac { 7 }{ x-y } \quad =10.$

Solution:

Concept: “If the number of equations given and the number of variables in the system of equations are equal, then there is an exact or unique solution exists.”          

Let $\frac{1}{(x+y)} = a \quad and\quad\frac{1}{(x-y)} = b;$

On substitution, the system of equations become

5a - 2b = -1                                                                         ---------------(1)

15a +7b = 10.                                                                         ---------------(2)

To eliminate a, multiply the (1) equation by factor 3, which gives:

15a - 6b = -3                                                                         ---------------(3)

Elimination method (Eliminating one variable and finding value of other variable).

(2) – (3) gives:

        (15a + 7b) - (15a – 6b) = 10-(-3)

        15a + 7b – 15a + 6b = 10 + 3

        13b = 13

         b = 1

Substitute the value of b in any of (1) or (3), which gives:

                                   $a = \frac{1}{5}.$

To find x and y:

From the assumption made, substitute a and b values:

$\frac{1}{(x+y)} = aand\frac{1}{(x-y)} = b.$

$\frac{1}{(x+y)} = \frac{1}{5};$

$\frac{1}{(x-y)} =\frac{1}{1};$

x + y = 5; x – y = 1

By adding the above equations:

(x + y) + (x-y) = 5+1

       2x = 6

         x = 3.

On substitution,

3 + y = 5

     y = 2

values are x = 3 and y = 2;"