Question
Solve :$11(x - 5) + 10(y - 2) + 54 = 0;7(2x - 1) + 9(3y - 1) = 25$
✓
Answer
$11( x - 5 ) + 10( y - 2 ) + 54 = 0($given$)$
$\Rightarrow 11x - 55 + 10y - 20 + 54 = 0$
$\Rightarrow 11x + 10y - 21 = 0$
$\Rightarrow 11x + 10y = 21\dots....(1)$
$7( 2x - 1 ) + 9(3y - 1) = 25($given$)$
$\Rightarrow 14x - 7 + 27y - 9 = 25$
$\Rightarrow 14x + 27y - 16 = 25$
$\Rightarrow 14x + 27y = 41\dots.....(2)$ Multiplying equation $(1)$ by $27$ and equation $(2)$ by $10,$ we get,
$297x + 270y = 567\dots....(3)$
$140x + 270y = 410\dots.....(4)$ Subtracting equation $(4)$ from equation $(3)$, we get
$157x = 157$
$\Rightarrow x = 1$ Substituting $x = 1$ in equation $(1),$ we get,
$11 x 1 + 10y = 21$
$\Rightarrow 10y = 10$
$\Rightarrow y = 1$
$\therefore $ Solution set is $x = 1$ and $y = 1.$
$\Rightarrow 11x - 55 + 10y - 20 + 54 = 0$
$\Rightarrow 11x + 10y - 21 = 0$
$\Rightarrow 11x + 10y = 21\dots....(1)$
$7( 2x - 1 ) + 9(3y - 1) = 25($given$)$
$\Rightarrow 14x - 7 + 27y - 9 = 25$
$\Rightarrow 14x + 27y - 16 = 25$
$\Rightarrow 14x + 27y = 41\dots.....(2)$ Multiplying equation $(1)$ by $27$ and equation $(2)$ by $10,$ we get,
$297x + 270y = 567\dots....(3)$
$140x + 270y = 410\dots.....(4)$ Subtracting equation $(4)$ from equation $(3)$, we get
$157x = 157$
$\Rightarrow x = 1$ Substituting $x = 1$ in equation $(1),$ we get,
$11 x 1 + 10y = 21$
$\Rightarrow 10y = 10$
$\Rightarrow y = 1$
$\therefore $ Solution set is $x = 1$ and $y = 1.$
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