How do you solve # 3/(5(x-2)) = x-4#?

1 Answer
EZ as pi
Jul 23, 2016

#x = 4.265 " or "x = 1.735#

Explanation:

Write the equation as #3/(5(x-2)) = (x-4)/1#

There is now one fraction term on each side and we can cross-multiply. We would have had the same result if we had multiplied both sides by the denominator.

#5(x-2)(x-4) = 3#

#5(x^2-4x-2x+8)-3=0" make a quadratic =0"#

#5x^2-20x -10x+40-3=0#

#5x^2-30x+37 = 0#

THere are no factors of 5 and 37 which will make 30, we need to use the formula: #a=5, b=-30 c=37#

#x = (-b +-sqrt(b^2-4ac))/(2a)#

#x = (-(-30) +-sqrt((-30)^2-4(5)(37)))/(2(5))#

#x = (30+-sqrt(900-740))/10#

#x = (30+sqrt160)/10" or "x = (30-sqrt160)/10#

#x = 4.265 " or "x = 1.735#

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