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Mathematics 20 Online
OpenStudy (wintersuntime):

Can someone be nice enough to help me with this problem please -> 3/4x+1/2=3/2

OpenStudy (wintersuntime):

x=4/3

OpenStudy (jdoe0001):

hint: multiply both sides by "4x"

OpenStudy (wintersuntime):

they are all fractions

OpenStudy (wintersuntime):

@sebaxtiangb

OpenStudy (jdoe0001):

right... thus multiplying them by 4x, removes the denominator and you'd end up with numerators

OpenStudy (jdoe0001):

\(\bf \cancel{4x}\cdot \cfrac{3}{\cancel{4x}}+\cancel{4} x\cdot \cfrac{1}{\cancel{2}}=\cancel{4}x\cdot \cfrac{3}{\cancel{2}}\implies ?\)

OpenStudy (anonymous):

Subtract 1/2 on both sides first.

OpenStudy (jdoe0001):

\(\bf \cancel{4x}\cdot \cfrac{3}{\cancel{4x}}+\cancel{4} x\cdot \cfrac{1}{\cancel{2}}=\cancel{4}x\cdot \cfrac{3}{\cancel{2}}\implies 3+2x\cdot 1=2x\cdot 3 \\ \quad \\ 3+2x=6x\implies ?\)

OpenStudy (anonymous):

|dw:1438733335459:dw| Then subtract the terms on the right sides. You can even simplify it.

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