Question

If [(x²+3x-5)/3x-5] = [(x²+2x-7)/2x-7)]. Find the value of x

Answer 1

Answer:

Step-by-step explanation:

$\frac{x^{2}+3x-5 }{3x-5}$ = $\frac{x^{2}+2x-7 }{2x-7}$

(x²+3x-5) (2x-7) = (x²+2x-7)(3x-5)

2x(x²+3x-5)-7(x²+3x-5) = 3x(x²+2x-7)-5(x²+2x-7)

2x³+6x²-10x-7x²-21x+35 = 3x³+6x²-21x-5x²-10x+35

3x³-2x³+6x²-5x²-6x²+7x²-21x-10x+10x+21x+35-35=0

x³+2x²=0

x²(x+2)=0

Either,

x²=0

x = 0

Or,

x+2 =0

x= -2

Answer 2

Answer:

[(x²+3x-5)/3x-5] = [(x²+2x-7)/2x-7)]

We can write [(x²+3x-5)/3x-5] = [(x²+2x-7)/2x-7)] as

[(x²/3x-5)+(3x-5/3x-5)] = [(x²/2x-7)+(2x-7/2x-7)]

∴ [(x²/3x-5)+1] = [(x²/2x-7)+1]

∴ (x²/3x-5)+1 = (x²/2x-7)+1

Now, we have to cancel '1' of RHS from '1' of LHS.

∴ (x²/3x-5) = (x²/2x-7)

∴ x² = (x²/2x-7)*3x-5

∴ x² = [{3x-5(x²)}/2x-7]

∴ x² = $\sqrt{\frac{3x^{3} -5x^{2} }{2x-7} }$

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