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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
_ _ _
=
I have been tired since typing from a long time rest of the question you can solve
Question:
Solve the following simultaneous linear equations:
[tex]\sf\:\dfrac{27}{x\:-\:2}\:+\:\dfrac{31}{y\:+\:3}\:=\:85\:\&\\\\\sf\:\dfrac{31}{x\:-\:2}\:+\:\dfrac{27}{y\:+\:3}\:=\:89[/tex]
Answer:
The solution of the given simultaneous equations is
[tex]\boxed{\red{\sf\:(\:x\:,\:y\:)\:=\:\bigg(\:\dfrac{5}{2}\:,\:-\:2\:\bigg)}}[/tex]
Step-by-step-explanation:
The given simultaneous equations are
[tex]\sf\:\dfrac{27}{x\:-\:2}\:+\:\dfrac{31}{y\:+\:3}\:=\:85\:\:-\:-\:(\:1\:)\:\&\\\\\sf\:\dfrac{31}{x\:-\:2}\:+\:\dfrac{27}{y\:+\:3}\:=\:89\:\:\:-\:-\:(\:2\:)[/tex]
By substituting a for [tex]\sf\:\dfrac{1}{x\:-\:2}[/tex] and b for [tex]\sf\:\dfrac{1}{y\:+\:3}[/tex], 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,
[tex]\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}}}[/tex]
Now, again by substituting the value of b, we get,
[tex]\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}}[/tex]
The solution of the given simultaneous equations is
[tex]\boxed{\red{\sf\:(\:x\:,\:y\:)\:=\:\bigg(\:\dfrac{5}{2}\:,\:-\:2\:\bigg)}}[/tex]