Question

solve:x-1/x+2=2x-5/3x-7​

Answer 1

Answer:

as we have given the equation =>

=>\frac{(x-1) }{(x+1) }=\frac{(2x-5) }{(3x-7) }=>(x+1(x−1)=(3x−7)(2x−5)

cross multiply both sides , we get =>

=>(X-1) (3x-7) =(2x-5) (x+1)=>(X−1)(3x−7)=(2x−5)(x+1)

=>3x^{2}-7x-3x+7\: =\: 2x^{2}-3x-12=>3x 2−7x−3x+7=2x−3x−12

=>3x^{2}-10x+7\: =\: 2x^{2}-3x-12=>3x 2 −10x+7=2x 2−3x−12

=>3x^{2}-2x^{2}-10x+3x+7+5=0=>3x 2 −2x 2 −10x+3x+7+5=0

=>x^{2}-7x+12\:=\:0=>x 2 −7x+12=0

=>x^{2}-3x-4x+12=0=>x 2 −3x−4x+12=0

=>x(x-3) \: -\:4(x-3) =\: 0=>x(x−3)−4(x−3)=0

=>(x-3) (x-4) =0=>(x−3)(x−4)=0

=>(x-3) =0 or(x-4) =0=>(x−3)=0or(x−4)=0

=>x = 3 or x= 4=>x=3 or x=4

our factors for given equation are 3 & 4.

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Answer 2

Answer:

Step-by-step explanation:

The equation:$7^(2x+1) - 7^(x+1) = -7$can be written as

$7*7^2x -2*7*7^x + 7 = 0,$ divide by 7 to get

$7^2x- 2*7^x +1 = 0. Let 7^x = p,$ then

$p^2 - 2p + 1 = 0,$ or

$(p-1)^2 = 0,$ or

$p-1 = 0$, or

$p = 1,$ or

$7^x = 1$ which means$x= 0.$

hope it helps

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