if the numerator and denominator of a fraction 3 by 4 in it be increased by a certain number and dose of friction 5 by 6 be decreased by the same number the result will be same find the number
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given fraction = 3/4 and 5/6
let those be increased and decreased by x
(3+x)/(4+x) = (5-x)/(6-x)
(3+x)(6-x) = (5-x)(4+x)
18 -3x +6x -x^2 = 20- 5x -4x -x^2
18 +3x = 20+ x
2x = 2
x=1
hope that helped you
please mark my answer as brainliest
given fraction = 3/4 and 5/6
let those be increased and decreased by x
(3+x)/(4+x) = (5-x)/(6-x)
(3+x)(6-x) = (5-x)(4+x)
18 -3x +6x -x^2 = 20- 5x -4x -x^2
18 +3x = 20+ x
2x = 2
x=1
hope that helped you
please mark my answer as brainliest
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