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Let x and y be non-zero numbers. If 2x-3y=0, then (x+3y):(x+2y)=?
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\[2x=3y\implies \frac{x}{y}=\frac{3}{2}\] x+3y:x+2y means \[\frac{ x+3y }{ x+2y }\] now divide each term by y in this fraction
(x/y + 2y)/(x/y + y)
\[\large\frac{ x+3y }{ x+2y }=\frac{ \frac{x}{y}+\frac{3y}{y} }{\frac{x}{y}+\frac{2y}{y} }=\frac{\frac{x}{y}+3}{\frac{x}{y}+2}\]
now substitute \[\frac{x}{y}=\frac{3}{2}\]
ohh, i miss it
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you dont understand?
9/7? i understand, just calculation mistake
9:7
okay alternatively we cud have put \[x=\frac{3y}{2}\] then substitute in the equation \[\frac{ x+3y }{ x+2y }=\frac{ \frac{ 3y }{ 2 } +3y}{ \frac{ 3y }{ 2 }+2y}=\frac{ \frac{ 9 }{ 2 }y }{ \frac{ 7 }{ 2 }y }=\frac{ 9 }{ 7 }\]
ohh i see, thank you
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