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Question
If `[ 2x - 3y ]/[ 3z + y] = [ z - y ]/[ z - x ] = [ x + 3z ]/[ 2y - 3x]` then prove that every ratio = `x/y`.
Solution
`[ 2x - 3y ]/[ 3z + y] = [ z - y ]/[ z - x ] = [ x + 3z ]/[ 2y - 3x]`
⇒ `[ 2x - 3y]/[ 3z + y] = [3( z - y)]/[ 3( z - x)] = [ x + 3z ]/[ 2y - 3x]` ...by theorem of equal ratios,
⇒ `{2x - 3y - 3( z - y) + x + 3z }/{ 3z + y - 3( z - x) + 2y - 3x }`
⇒ `{2x - 3y - 3z + 3y + x + 3z }/{ 3z + y - 3z + 3x + 2y - 3x }`
⇒ `(3x)/(3y)`
⇒ `x / y`
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