Question

I have three expression questions I need help solving for 1. x(25+2x-y)___________ -ZX= -3, Y= 4 Z= -1 2.2(x-3y)________ 6X= 6, Y= 1 3.(x^2+5)(x^2-x+2)-3X= -4

Answer 1

The first expression can be evaluated as follows:

$\frac{x(25+2x-y)}{-z};x=-3,y=4,z=-1$

Substitute the given values to get:

$\begin{gathered} \frac{x(25+2x-y)}{-z}=\frac{-3(25+2(-3)-4)}{-(-1)} \\ =\frac{-3(25-6-4)}{1} \\ =\frac{-3(25-10)}{1} \\ =-3(15) \\ =-45 \end{gathered}$

The second expression can be evaluated as follows:

$(3x^2)\frac{2(x-3y)}{6};x=6,y=1$

Substitute the given values to get:

$\begin{gathered} (3(6^2))\frac{2(6-3(1))}{6}=\frac{2(3)}{6} \\ =(3\times36)\frac{6}{6} \\ =108 \end{gathered}$

The third expression can be evaluated as follows:

$(x^2+5)(x^2-x+2)-3;x=-4$

Substitute the given value to get:

$\begin{gathered} \lbrack(-4)^2+5\rbrack\lbrack(-4)^2-(-4)+2\rbrack-3=(16+5)(16+4+2)-3 \\ =(21)(22)-3 \\ =462-3 \\ =459 \end{gathered}$