Question

Use the given values to complete each table (sorry the image is tipped)

Answer 1

(2) The solutions are
`-10,-2,6`

(3) The solutions are
`20,(5)/(2) ,32`

Step-by-step explanation:

(2) The value of the equation
`2z-a` can be found by substituting the values for z and a.

For
`z=-4` and
`a=2`, we get,

`\begin{aligned}2 z-a &=2(-4)-2 \\&=-8-2 \\&=-10\end{aligned}`

For
`z=0` and
`a=2`, we get,

`\begin{aligned}2 z-a &=2(0)-2 \\&=0-2 \\&=-2\end{aligned}`

For
`z=4` and
`a=2`, we get,

`\begin{aligned}2 z-a &=2(4)-2 \\&=8-2 \\&=6\end{aligned}`

Thus, the solutions are
`-10,-2,6`

(3) The value of the equation
`10 x^(2) /(y+1)` can be found by substituting the values for x and y.

For
`x=2` and
`y=1`, we get,

`\begin{aligned}10 x^(2) /(y+1) &=10(2)^(2) /(1+1) \\&=10(4) /(2) \\&=40 / 2 \\&=20\end{aligned}`

For
`x=-1` and
`y=3`, we get,

`\begin{aligned}10 x^(2) /(y+1) &=10(-1)^(2) /(3+1) \\&=10(1) /(4) \\&=10 / 4 \\&=(5)/(2) \end{aligned}`

For
`x=-4` and
`y=4`, we get,

`\begin{aligned}10 x^(2) /(y+1) &=10(-4)^(2) /(4+1) \\&=10(16) /(5) \\&=160 / 5 \\&=32 \end{aligned}`

Thus, the solutions are
`20,(5)/(2) ,32`