(3-2x)/5=(4-3x)/7

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Solution for (3-2x)/5=(4-3x)/7 equation: 



(3-2x)/5=(4-3x)/7

We move all terms to the left:

(3-2x)/5-((4-3x)/7)=0

We add all the numbers together, and all the variables

(-2x+3)/5-((-3x+4)/7)=0

We calculate fractions


(-2x)/()+(-((-3x+4)*5)/()=0



We calculate terms in parentheses: +(-((-3x+4)*5)/(), so:

-((-3x+4)*5)/(

We multiply all the terms by the denominator


-((-3x+4)*5)



We calculate terms in parentheses: -((-3x+4)*5), so:

(-3x+4)*5

We multiply parentheses

-15x+20

Back to the equation:

-(-15x+20)



We get rid of parentheses

15x-20

Back to the equation:

+(15x-20)



We get rid of parentheses

(-2x)/()+15x-20=0

We multiply all the terms by the denominator


(-2x)+15x*()-20*()=0

We add all the numbers together, and all the variables

(-2x)+15x*()=0

We get rid of parentheses

-2x+15x*()=0

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