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(A) First,simplify the given equations:$1$) $5(x-2) + 3y = 1 \implies 5x - 10 + 3y = 1 \implies 5x + 3y = 11$ (Equation $1$)$2$) $3(x+1) + 5(y-3) = 1

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$(1, 2)$

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(A) First,simplify the given equations:$1$) $5(x-2) + 3y = 1 \implies 5x - 10 + 3y = 1 \implies 5x + 3y = 11$ (Equation $1$)$2$) $3(x+1) + 5(y-3) = 1 \implies 3x + 3 + 5y - 15 = 1 \implies 3x + 5y = 13$ (Equation $2$)To eliminate $y$,multiply Equation $1$ by $5$ and Equation $2$ by $3$:$25x + 15y = 55$ (Equation $3$)$9x + 15y = 39$ (Equation $4$)Subtract Equation $4$ from Equation $3$:$(25x - 9x) + (15y - 15y) = 55 - 39$$16x = 16 \implies x = 1$Substitute $x = 1$ into Equation $1$:$5(1) + 3y = 11$$5 + 3y = 11$$3y = 6 \implies y = 2$Thus,the solution is $(x, y) = (1, 2)$.

Solve the following pairs of equations by the method of elimination:
$5(x-2) + 3y = 1$
$3(x+1) + 5(y-3) = 1$

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Solve the following pairs of equations by the method of elimination:$5(x-2) + 3y = 1$$3(x+1) + 5(y-3) = 1$