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08 Jan 2020, 08:54
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Hi all,
I am in the middle of a TTP question and seem to be getting stuck on particular step.
Original question stem for context
If \(x-5 = \frac{2}{3}(y-5)\), and \(x+10=\frac{5}{6}(y+10)\), what is the value of xy?
First step: [Equation one] eliminate the fraction by multiplying the entire equation by 3.
TTP says:
\(3(x-5=\frac{2}{3}(y-5))\)
\(3x-15=2(y-5)\)
\(3x-15=2y-10\)
\(3x=2y+5\)
But why don't you multiply the \((y-5) \) by 3 as well?
I was doing this:
\(3(x-5=\frac{2}{3}(y-5))\)
\(3x-15=2(3y-15)\)
\(3x-15=6y-30\)
\(3x=6y-15\)
\(x=2y-5\)
I am in the middle of a TTP question and seem to be getting stuck on particular step.
Original question stem for context
If \(x-5 = \frac{2}{3}(y-5)\), and \(x+10=\frac{5}{6}(y+10)\), what is the value of xy?
First step: [Equation one] eliminate the fraction by multiplying the entire equation by 3.
TTP says:
\(3(x-5=\frac{2}{3}(y-5))\)
\(3x-15=2(y-5)\)
\(3x-15=2y-10\)
\(3x=2y+5\)
But why don't you multiply the \((y-5) \) by 3 as well?
I was doing this:
\(3(x-5=\frac{2}{3}(y-5))\)
\(3x-15=2(3y-15)\)
\(3x-15=6y-30\)
\(3x=6y-15\)
\(x=2y-5\)
Abhi077
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08 Jan 2020, 11:07
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Here's why.
When you multiply both the equation by 3, it's like this 3 X (x - 5) = 3 X \(\frac{ 2}{3}\) ( y - 5), you forgot to cancel out 3. Hope this helps
When you multiply both the equation by 3, it's like this 3 X (x - 5) = 3 X \(\frac{ 2}{3}\) ( y - 5), you forgot to cancel out 3. Hope this helps
09 Jan 2020, 06:36
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Abhi077
Here's why.
When you multiply both the equation by 3, it's like this 3 X (x - 5) = 3 X \(\frac{ 2}{3}\) ( y - 5), you forgot to cancel out 3. Hope this helps
When you multiply both the equation by 3, it's like this 3 X (x - 5) = 3 X \(\frac{ 2}{3}\) ( y - 5), you forgot to cancel out 3. Hope this helps
Thanks for your response Abhi077 but why don't you distribute the 3 to (y-5)?
