Question

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Answer 1

Step-by-step explanation:

let assume that,

1/x-2 = m and 1/y+3 = n

27m + 31n = 85 ....(3)

31m + 27n = 89 ....(4)

add equation (3) , (4)

27m + 31n = 85

31m + 27 n = 89

=

58m + 58n = 174 ... (5)

subtract equation (3) , (4)

27m + 31n = 85

31m + 27n = 89

_ _ _

=

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Answer 2

Question:

Solve the following simultaneous linear equations:

$\sf\:\dfrac{27}{x\:-\:2}\:+\:\dfrac{31}{y\:+\:3}\:=\:85\:\&\\\\\sf\:\dfrac{31}{x\:-\:2}\:+\:\dfrac{27}{y\:+\:3}\:=\:89$

Answer:

The solution of the given simultaneous equations is

$\boxed{\red{\sf\:(\:x\:,\:y\:)\:=\:\bigg(\:\dfrac{5}{2}\:,\:-\:2\:\bigg)}}$

Step-by-step-explanation:

The given simultaneous equations are

$\sf\:\dfrac{27}{x\:-\:2}\:+\:\dfrac{31}{y\:+\:3}\:=\:85\:\:-\:-\:(\:1\:)\:\&\\\\\sf\:\dfrac{31}{x\:-\:2}\:+\:\dfrac{27}{y\:+\:3}\:=\:89\:\:\:-\:-\:(\:2\:)$

By substituting a for $\sf\:\dfrac{1}{x\:-\:2}$ and b for $\sf\:\dfrac{1}{y\:+\:3}$, we get,

27a + 31b = 85 - - ( 3 )

31a + 27b = 89 - - ( 4 )

By adding equations ( 3 ) and ( 4 ), we get,

27a + 31a + 31b + 27b = 85 + 89

⇒ 58a + 58b = 174

⇒ a + b = 3 - - - ( 5 ) [ Dividing by 58 ]

Now, by subtracting equation ( 3 ) from equation ( 4 ), we get,

⇒ 31a - 27a + 27b - 31b = 89 - 85

⇒ 4a - 4b = 4

⇒ a - b = 1 - - - ( 6 ) [ Dividing by 4 ]

Now, adding equations ( 5 ) & ( 6 ), we get,

⇒ a + a + b - b = 3 + 1

⇒ 2a + 0 = 4

⇒ 2a = 4

⇒ a = 4 ÷ 2

⇒ a = 2

By substituting a = 2 in equation ( 5 ), we get,

⇒ a + b = 3 - - ( 5 )

⇒ 2 + b = 3

⇒ b = 3 - 2

⇒ b = 1

Now, again by substituting the value of a, we get,

$\sf\:a\:=\:\dfrac{1}{x\:-\:2}\\\\\\\implies\sf\:\dfrac{1}{x\:-\:2}\:=\:2\\\\\\\implies\sf\:1\:=\:2\:(\:x\:-\:2\:)\\\\\\\implies\sf\:1\:=\:2x\:-\:4\\\\\\\implies\sf\:2x\:=\:1\:+\:4\\\\\\\implies\sf\:2x\:=\:5\\\\\\\implies\boxed{\red{\sf\:x\:=\:\dfrac{5}{2}}}$

Now, again by substituting the value of b, we get,

$\sf\:b\:=\:\dfrac{1}{y\:+\:3}\\\\\\\implies\sf\:\dfrac{1}{y\:+\:3}\:=\:1\\\\\\\implies\sf\:1\:=\:y\:+\:3\\\\\\\implies\sf\:y\:=\:1\:-\:3\\\\\\\implies\boxed{\red{\sf\:y\:=\:-\:2}}$

The solution of the given simultaneous equations is

$\boxed{\red{\sf\:(\:x\:,\:y\:)\:=\:\bigg(\:\dfrac{5}{2}\:,\:-\:2\:\bigg)}}$