Math, asked by nimmalasrinivas17, 1 month ago

3 x -5 y = -1 and x –y = -1 Solve the pair of equations by substitution method.

Answers

Answered by ISAlishaTripathy
4

Answer:

x=-2 and y=-1

Step-by-step explanation:

The explanation is in the above attachment.

Attachments:
Answered by DüllStâr
38

 \bf \pink{Solution:}

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 \sf3x  - 5y = -1... ... ... ... ... ... ... \boxed{ \sf1}

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 \sf{}x - y =  - 1 \: ... ... ... ... ... ... ... \boxed{ \sf2}

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Solving equation 1 :

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:  \implies \sf3x  - 5y = -1

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:  \implies \sf3x = -1 + 5y

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:  \implies \sf3x  = 5y - 1

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:  \implies \sf{}x  = \dfrac{5y - 1}{3}... ... ... ... \boxed{ \sf1}

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Now put the value of x in equation 2:

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 :  \implies \sf{}x - y =  - 1

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 :  \implies \sf{} \dfrac{5y - 1}{3}  - y =  - 1

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 :  \implies \sf{} \dfrac{5y - 1 - (y \times 3)}{3} =  - 1

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 :  \implies \sf{} \dfrac{5y - 1 -3y}{3} =  - 1

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 :  \implies \sf{} \dfrac{5y -3y - 1}{3} =  - 1

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 :  \implies \sf{} \dfrac{2y - 1}{3} =  - 1

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 :  \implies \sf{}2y - 1=  - 1 \times 3

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 :  \implies \sf{}2y - 1=  -3

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 :  \implies \sf{}2y =  -3 + 1

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 :  \implies \sf{}2y =  -2

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 :  \implies \sf{}y =  \dfrac{ - 2}{2}

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 :  \implies \sf{}y =  \dfrac{ - \cancel2}{\cancel2}

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 :  \implies \sf{}y =  \dfrac{ -1}{1}

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 :  \implies \sf{}y = - 1

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 \therefore \underline{ \sf{}value \: of \: y \: is \:  \textbf{ \textsf{ - 1}}}

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━━━━━━━━━━━━━━━

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Now Let's find value of x

put value of y in Equation 1.

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:  \implies \sf{}x  = \dfrac{5 \times  - 1 -  1}{3}

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:  \implies \sf{}x  = \dfrac{ - 5 - 1}{3}

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:  \implies \sf{}x  = \dfrac{ - 6}{3}

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:  \implies \sf{}x  = \dfrac{ - (2 \times 3)}{3}

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:  \implies \sf{}x  = \dfrac{ - (2 \times \cancel3)}{\cancel3}

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:  \implies \sf{}x  = \dfrac{ - (2 \times 1)}{1}

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:  \implies \sf{}x  =  - 2 \times 1

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:  \implies \sf{}x  =  - 2

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 \therefore \underline{ \sf{}value \: of \: x\: is \:  \textbf{ \textsf{ - 2}}}

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━━━━━━━━━━━━━━━

Verification:

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:  \implies \sf{}x-y=-1

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put value of x in this equation.

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:  \implies \sf{-2-(-1)=-1}

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:  \implies \sf{}-2+1=-1

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:  \implies \sf{}-1=-1

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 \bf LHS = RHS

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 \bf Hence\:Verified!

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