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Solve the following system of equations algebraically :
$30x+44y = 10$; $40x+55y = 13$
$30x+44y = 10$; $40x+55y = 13$
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Given equations can be rewritten as
$120 x + 176 y = 40$ ...(i)
$120 x + 165 y = 39$ ...(ii)
Subtracting to get $y = \frac{1}{11}$
Substituting to get $x = \frac{1}{5}$
$120 x + 176 y = 40$ ...(i)
$120 x + 165 y = 39$ ...(ii)
Subtracting to get $y = \frac{1}{11}$
Substituting to get $x = \frac{1}{5}$