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5y^-2 + 1= 6y^-1 how do you solve it by step?
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y=1,5
I mean step
plug in the value
little lost on the "plug in the value" can you give me a example?
Take this 5y^-2 and 6y^-1 Subtract it
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\[\frac{5}{y^2}+1=\frac{6}{y}\] is how i read it. is that wrong?
yea
sorry my computer is acting up
ok so maybe \[1=\frac{6}{y}-\frac{5}{y^2}\] \[1=\frac{6y-5}{y^2}\] \[y^2=6y-5\] \[y^2-6y+5=0\] \[(y-5)(y-1)=0\] \[y=5, y=1\]
That the part what I'm missing thanks
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