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Mathematics 7 Online
OpenStudy (anonymous):

Let x and y be non-zero numbers. If 2x-3y=0, then (x+3y):(x+2y)=?

OpenStudy (anonymous):

\[2x=3y\implies \frac{x}{y}=\frac{3}{2}\] x+3y:x+2y means \[\frac{ x+3y }{ x+2y }\] now divide each term by y in this fraction

OpenStudy (anonymous):

(x/y + 2y)/(x/y + y)

OpenStudy (anonymous):

\[\large\frac{ x+3y }{ x+2y }=\frac{ \frac{x}{y}+\frac{3y}{y} }{\frac{x}{y}+\frac{2y}{y} }=\frac{\frac{x}{y}+3}{\frac{x}{y}+2}\]

OpenStudy (anonymous):

now substitute \[\frac{x}{y}=\frac{3}{2}\]

OpenStudy (anonymous):

ohh, i miss it

OpenStudy (anonymous):

you dont understand?

OpenStudy (anonymous):

9/7? i understand, just calculation mistake

OpenStudy (anonymous):

9:7

OpenStudy (anonymous):

okay alternatively we cud have put \[x=\frac{3y}{2}\] then substitute in the equation \[\frac{ x+3y }{ x+2y }=\frac{ \frac{ 3y }{ 2 } +3y}{ \frac{ 3y }{ 2 }+2y}=\frac{ \frac{ 9 }{ 2 }y }{ \frac{ 7 }{ 2 }y }=\frac{ 9 }{ 7 }\]

OpenStudy (anonymous):

ohh i see, thank you

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