Question

15
22
E = 5
21.
x-y x+y
40 55
+ = 13,
x-y x+y
X#y, x#-y​

Answer 1

Answer:

unable to understand your question.

Sorry.

Answer 2

Step-by-step explanation:

Answer:

\begin{lgathered}x = 8 \\y= 3\end{lgathered}

x=8

y=3

Step-by-step explanation:

\begin{lgathered}Given \:pairs \:of\: equation \\for\:x\:and\:y;\\\frac{15}{x-y}+\frac{22}{x+y}=5--(1)\end{lgathered}

Givenpairsofequation

forxandy;

x−y

15

+

x+y

22

=5−−(1)

\frac{40}{x-y}+\frac{55}{x+y}=13--(2)

x−y

40

+

x+y

55

=13−−(2)

\begin{lgathered}Taking \:\frac{1}{x-y}=m\:and\\\frac{1}{x+y}=n,\:the \\system \:of\: equations\:becomes \\15m+22n=5---(3),\\40m+55n=13---(4)\end{lgathered}

Taking

x−y

1

=mand

x+y

1

=n,the

systemofequationsbecomes

15m+22n=5−−−(3),

40m+55n=13−−−(4)

Now, multiply equation (3) by 5 ,and equation (4) by 2 ,we get

\begin{lgathered}75m+110n=25--(5)\\80m+110n=26---(6)\end{lgathered}

75m+110n=25−−(5)

80m+110n=26−−−(6)

Subtract equation (5) from equation (6), we get

\implies 5m = 1⟹5m=1

\implies m = \frac{1}{5}⟹m=

5

1

Substitute\: m = \frac{1}{5}\:in \: equation \: (3), \: we\:getSubstitutem=

5

1

inequation(3),weget

3+22n= 53+22n=5

\implies 22n = 5-3⟹22n=5−3

\implies 22n = 2⟹22n=2

\implies n = \frac{2}{22}⟹n=

22

2

\implies n = \frac{1}{11}⟹n=

11

1

\begin{lgathered}m = \frac{1}{x-y}=\frac{1}{5}\\x-y = 5 ---(7)\end{lgathered}

m=

x−y

1

=

5

1

x−y=5−−−(7)

\begin{lgathered}n = \frac{1}{x+y}=\frac{1}{11}\\x+y =11 ---(8)\end{lgathered}

n=

x+y

1

=

11

1

x+y=11−−−(8)

Add equation (7) and (8), we get

\implies 2x=16⟹2x=16

\implies x = 8⟹x=8

Substitute \: x=8 \: in \: (8)\:,we\: getSubstitutex=8in(8),weget

\implies 8+y=11⟹8+y=11

\implies y = 11-8 = 3⟹y=11−8=3

Therefore,

\begin{lgathered}x = 8 \\y= 3\end{lgathered}

x=8

y=3