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Mathematics

Using properties of proportion, find the value of ‘x’:

6x2+3x−53x−5=9x2+2x+52x+5  x≠0\dfrac{6x^{2} + 3x - 5}{3x - 5} = \dfrac{9x^{2} + 2x + 5}{2x + 5} \; x \neq 0

Ratio Proportion

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Answer

Given,

⇒6x2+3x−53x−5=9x2+2x+52x+5\Rightarrow \dfrac{6x^{2} + 3x - 5}{3x - 5} = \dfrac{9x^{2} + 2x + 5}{2x + 5}

Applying componendo and dividendo,

⇒6x2+3x−5+(3x−5)6x2+3x−5−(3x−5)=9x2+2x+5+(2x+5)9x2+2x+5−(2x+5)⇒6x2+3x−5+3x−56x2+3x−5−3x+5=9x2+2x+5+2x+59x2+2x+5−2x−5⇒6x2+6x−106x2=9x2+4x+109x2⇒6x2+6x−106=9x2+4x+109⇒9×(6x2+6x−10)=6×(9x2+4x+10)⇒54x2+54x−90=54x2+24x+60⇒54x2+54x−90−54x2−24x−60=0⇒54x2−54x2+54x−24x−90−60=0⇒30x−150=0⇒30x=150⇒x=15030⇒x=5.\Rightarrow \dfrac{6x^{2} + 3x - 5 + (3x - 5)}{6x^{2} + 3x - 5 -(3x - 5)} = \dfrac{9x^{2} + 2x + 5 + (2x + 5)}{9x^{2} + 2x + 5 - (2x + 5)} \\[1em] \Rightarrow \dfrac{6x^{2} + 3x - 5 + 3x - 5}{6x^{2} + 3x - 5 - 3x + 5} = \dfrac{9x^{2} + 2x + 5 + 2x + 5}{9x^{2} + 2x + 5 - 2x - 5} \\[1em] \Rightarrow \dfrac{6x^{2} + 6x - 10}{6x^{2}} = \dfrac{9x^{2} + 4x + 10}{9x^{2}} \\[1em] \Rightarrow \dfrac{6x^{2} + 6x - 10}{6} = \dfrac{9x^{2} + 4x + 10}{9} \\[1em] \Rightarrow 9 \times (6x^{2} + 6x - 10) = 6 \times (9x^{2} + 4x + 10) \\[1em] \Rightarrow 54x^{2} + 54x - 90 = 54x^{2} + 24x + 60 \\[1em] \Rightarrow 54x^{2} + 54x - 90 - 54x^{2} - 24x - 60 = 0 \\[1em] \Rightarrow 54x^{2} - 54x^{2} + 54x - 24x - 90 - 60 = 0 \\[1em] \Rightarrow 30x - 150 = 0 \\[1em] \Rightarrow 30x = 150 \\[1em] \Rightarrow x = \dfrac{150}{30} \\[1em] \Rightarrow x = 5.

Hence, x = 5.

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