Question

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

Answer:

x=-2 and y=-1

Step-by-step explanation:

The explanation is in the above attachment.

Answer 2

$\bf \pink{Solution:}$

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

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

Solving equation 1 :

$: \implies \sf3x - 5y = -1$

$: \implies \sf3x = -1 + 5y$

$: \implies \sf3x = 5y - 1$

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

Now put the value of x in equation 2:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Now Let's find value of x

put value of y in Equation 1.

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

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

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

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

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

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

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

$: \implies \sf{}x = - 2$

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

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Verification:

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

put value of x in this equation.

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

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

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

$\bf LHS = RHS$

$\bf Hence\:Verified!$