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Mathematics 11 Online
OpenStudy (anonymous):

5y^-2 + 1= 6y^-1 how do you solve it by step?

OpenStudy (anonymous):

y=1,5

OpenStudy (anonymous):

I mean step

OpenStudy (anonymous):

plug in the value

OpenStudy (anonymous):

little lost on the "plug in the value" can you give me a example?

OpenStudy (anonymous):

Take this 5y^-2 and 6y^-1 Subtract it

OpenStudy (anonymous):

\[\frac{5}{y^2}+1=\frac{6}{y}\] is how i read it. is that wrong?

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

sorry my computer is acting up

OpenStudy (anonymous):

ok so maybe \[1=\frac{6}{y}-\frac{5}{y^2}\] \[1=\frac{6y-5}{y^2}\] \[y^2=6y-5\] \[y^2-6y+5=0\] \[(y-5)(y-1)=0\] \[y=5, y=1\]

OpenStudy (anonymous):

That the part what I'm missing thanks

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