Question

5 upon X + Y + 1 upon x minus y = 2 to 15 upon X + Y - 5 upon x minus y = - 2 ​

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• A linear Equations
• $\tt\dfrac{5}{X+Y}+\dfrac{1}{X-Y}=2$
• &
• $\tt\dfrac{15}{X+Y}-\dfrac{5}{X-Y}=2$

${\sf{\green{\underline{\large{To\:find}}}}}$

${\sf{\pink{\underline{\Large{Explanation}}}}}$

☞

$\tt\pink{\dfrac{5}{X+Y}+\dfrac{1}{X-Y}=2\rightarrow{(1)}}$

&

$\tt\green{\dfrac{15}{X+Y}-\dfrac{5}{X-Y}=-2\rightarrow{(2)}}$

Solving By elimination method

Let 1/(x-y) = a

and 1(X+y)=b

Now new equation will be

5b+a=2________(3)

15b-5a=-2______(4)

Multiplying by 3 in equation (3)

(5b+a)3=2×3

=>15b+3a=6

Now By Elimination

15b-5a=-2

15b+3a=6

_______

=>-8a=-8

=>a=1

¶utting the value of a in equation 3

5b+a=2

=>5b+1=2

=>b=1/5

Now substituting the value

1/(x-y) = a

and 1(X+y)=b

=>1/(x-y) = 1. & 1(X+y)=1/5

=>x-y=1 _____(5)

=>X+y=5_____(6)

______

2x=6

=>X=3

¶utting the value of X in equation(6)

=>3+y=5

=>y=2

Answer 2

Answer:

X=3 AND Y=2 IS THE ANSWER!

Step-by-step explanation:

A linear Equations

&

☞

&

Solving By elimination method

Let 1/(x-y) = a

and 1(X+y)=b

Now new equation will be

5b+a=2________(3)

15b-5a=-2______(4)

Multiplying by 3 in equation (3)

(5b+a)3=2×3

=>15b+3a=6

Now By Elimination

15b-5a=-2

15b+3a=6

_______

=>-8a=-8

=>a=1

¶utting the value of a in equation 3

5b+a=2

=>5b+1=2

=>b=1/5

Now substituting the value

1/(x-y) = a

and 1(X+y)=b

=>1/(x-y) = 1. & 1(X+y)=1/5

=>x-y=1 _____(5)

=>X+y=5_____(6)

______

2x=6

=>X=3

¶utting the value of X in equation(6)

=>3+y=5

=>y=2