Respuesta :

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.

Happil

Evaluating Rational Expressions

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.

Solution of a:

[tex]\frac{2x -y}{y^{2} +1}\\ \frac{2(1) -(2)}{(2)^{2} +1} \\ \frac{2 -(2)}{4+1} \\ \frac{0}{5} \\ 0[/tex]

Answer of a:

[tex]0[/tex]

Solution of b:

[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]

Answer of b:

[tex]-\frac{3}{26}\\[/tex]

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