Exercise 47 Page P21

To find points that satisfy the equation $y=\frac{1}{2}x-5,$ we assign any value we want to the variable $x$ and solve the equation for $y.$ Let's begin by setting $x_1=0.$ Substituting it into the equation above, we will find the value of $y_1.$ The point $P(0,\text{-}5)$ satisfies the equation. For the remaining two points, let's summarize the computations in a table.

$x\text{-}$value	$y=\frac{1}{2}x-5$	$y\text{-}$value	Point
$\text{-}2$	$y=\frac{1}{2}(\text{-}2)-5$	$\text{-}6$	$Q(\text{-}2,\text{-}6)$
$2$	$y=\frac{1}{2}(2)-5$	$\text{-}4$	$R(2,\text{-}4)$

Now, let's graph these points.

Note that these points are just examples, and there are infinitely many other points that satisfy the given equation.