Will Give Brainliest if Correct
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Answer:
(a) 0
(b) -3/26
Step-by-step explanation:
(a)
We essentially plug 1 in for x and 2 in for y.
Our expression is [tex]\frac{2x-y}{y^2+1}[/tex], so making the correct substitutions yields:
[tex]\frac{2x-y}{y^2+1}=\frac{2*1-2}{2^2+1}=\frac{2-2}{4+1} =\frac{0}{5} =0[/tex]
So, the answer for part (a) is 0.
(b)
Again, we're now plugging in 1 for x and 5 for y:
[tex]\frac{2x-y}{y^2+1}=\frac{2*1-5}{5^2+1}=\frac{2-5}{25+1} =\frac{-3}{26} =-\frac{3}{26}[/tex]
So, the answer for part (b) is -3/26.
Step-by-step explanation:
So we have the rational expression [tex]\frac{2x -y}{y^{2} +1}\\[/tex]. All we have to do is to just plug the given values for what x and y are.
[tex]\frac{2x -y}{y^{2} +1}\\ \frac{2(1) -(2)}{(2)^{2} +1} \\ \frac{2 -(2)}{4+1} \\ \frac{0}{5} \\ 0[/tex]
[tex]0[/tex]
[tex]\frac{2x -y}{y^{2} +1}\\ \frac{2(1) -(5)}{(5)^{2} +1} \\ \frac{2 -(5)}{25+1} \\ \frac{-3}{26} \\ -\frac{3}{26}[/tex]
[tex]-\frac{3}{26}\\[/tex]