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(2x+5)(3x-2)=(2x-5)(3x-7)
We move all terms to the left:
(2x+5)(3x-2)-((2x-5)(3x-7))=0
We multiply parentheses ..
(+6x^2-4x+15x-10)-((2x-5)(3x-7))=0
We calculate terms in parentheses: -((2x-5)(3x-7)), so:We get rid of parentheses
(2x-5)(3x-7)
We multiply parentheses ..
(+6x^2-14x-15x+35)
We get rid of parentheses
6x^2-14x-15x+35
We add all the numbers together, and all the variables
6x^2-29x+35
Back to the equation:
-(6x^2-29x+35)
6x^2-6x^2-4x+15x+29x-10-35=0
We add all the numbers together, and all the variables
40x-45=0
We move all terms containing x to the left, all other terms to the right
40x=45
x=45/40
x=1+1/8
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