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Multiply both sides by \(x^2\). Or if you want to get rid of all fractions do \(30x^2\).
Just x^2?
Sure.
try it and see what happens.
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3x^2/(2x)(x^2) + 5x^2/(6x)(x^2)=4/(5x^2)
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3/(2x) +5/(6x)=4/(5x^2)
7/3x=4/5x^2
now what
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\[\frac{ 3 }{ 2x }+\frac{ 5 }{ 6x }=\frac{ 3*3+5 }{6x }=\frac{ 14 }{ 6x }=\frac{ 7 }{3x }\] \[\frac{ 7 }{ 3x }=\frac{ 4 }{5x^2 }\] cross multiply ,remember x not equal to zero.
35x=12?
\[35x^2=12x\] \[35 x^2-12x=0,x(35x-12)=0,x \neq 0,35x-12=0,x=\frac{ 12 }{ 35 }\]
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