Question

$\frac{ {x }^{2} - 3x + 7 }{3x - 7} = \frac{ {x}^{2} - 2x + 5 }{2x - 5}$
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Answer 1

Step-by-step explanation:

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Answer 2

ANSWER:

Given:

$\:\:\:\bullet\:\:\:\dfrac{x^2-3x+7}{3x-7}=\dfrac{x^2-2x+5}{2x-5}$

To Find:

• Value of x

Solution:

We are given that,

$:\longrightarrow\dfrac{x^2-3x+7}{3x-7}=\dfrac{x^2-2x+5}{2x-5}$

On cross-multiplying,

$:\implies(2x-5)(x^2-3x+7)=(3x-7)(x^2-2x+5)$

Solving the brackets,

$:\implies2x^3-6x^2+14x-5x^2+15x-35=3x^3-6x^2+15x-7x^2+14x-35$

On grouping like terms on each side separately,

$:\implies2x^3-(6x^2+5x^2)+(14x+15x)-35=3x^3-(6x^2+7x^2)+(15x+14x)-35$

Simplifying the terms,

$:\implies2x^3-11x^2+29x-35=3x^3-13x^2+29x-35$

Bringing all terms from LHS to RHS,

$:\implies0=3x^3-13x^2+29x-35-(2x^3-11x^2+29x-35)$

So,

$:\implies3x^3-13x^2+29x-35-2x^3+11x^2-29x+35=0$

Grouping like terms together,

$:\implies(3x^3-2x^3)-(13x^2-11x^2)+(29x-29x)+(35-35)=0$

Simplifying,

$:\implies x^3-2x^2+0+0=0$

So,

$:\implies x^3-2x^2=0$

Taking x² common,

$:\implies x^2(x-2)=0$

So,

$:\implies x^2=0\:\:and\:\:(x-2)=0$

Hence,

$:\implies\bf x=0,0\:\:and\:\:x=2$

Therefore, the value of x is 0, 0 and 2.