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प्रश्न
Using properties of proportion, find the value of ‘x’:
`(6x^2 + 3x − 5)/(3x − 5) = (9x^2 + 2x + 5)/(2x + 5); x ≠ 0`
बेरीज
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उत्तर
Given:
`(6x^2 + 3x − 5)/(3x − 5) = (9x^2 + 2x + 5)/(2x + 5); x ≠ 0`
`(6x^2 + 3x − 5 + 3x − 5)/(6x^2 + 3x − 5 − 3x + 5) = (9x^2 + 2x + 5 + 2x + 5)/(9x^2 + 2x + 5 − 2x − 5)`
Simplifying both sides:
`(6x^2 + 6x − 10)/(6x^2) = (9x^2 + 4x + 10)/(9x^2)`
Reducing further:
`(6x^2 + 6x − 10)/2 = (9x^2 + 4x + 10)/3`
Cross-multiplying:
18x2 + 18x − 30 = 18x2 + 8x + 20
10x = 50
x = 5
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