Chapter 3.1, Problem 10CYU

To determine

Graph each equation by making a table.

Answer to Problem 10CYU

This a graph parallel to x − axis.

Explanation of Solution

Given:

The equation: $-3+2y=-5$

Concept Used:

Simplify the equation:

  $\begin{array}{l}-3+2y=-5\\ -3+3+2y=-5+3\\ 2y=-2\\ \frac{2y}{2}=\frac{-2}{2}\\ y=-1\end{array}$ Equation: $y=-1$ is a line parallel to x − axis.

The equation of a line parallel to x-axis:

Let AB be a straight line parallel to x-axis at a distance b units from it. Then, clearly, all points on the line AB have the same x − value b. Thus, AB can be considered as the locus of a point at a distance b from x-axis and all points on the line AB satisfy the condition y = b.

Thus, if P(x, y) is any point on AB, then y = b.

Hence, the equation of a straight line parallel to x-axis at a distance b from it is y = b.

To graph the equation make y as subject and make a table by choosing random value of x and find the corresponding value of y.

Calculation:

Simplify the equation:

  $\begin{array}{l}-3+2y=-5\\ -3+3+2y=-5+3\\ 2y=-2\\ \frac{2y}{2}=\frac{-2}{2}\\ y=-1\end{array}$ Equation: $y=-1$ is a line parallel to x − axis.

Step 1: in this simplified equation $y=-1$ ; x is not present. We can rewrite the equation as $y=0·x-1$

Take random value of x and find the value of corresponding y

Make a table:

x	$y=0·x-1$	(x, y)
– 8	$\begin{array}{l}y=0·x-1\\ y=0·(-8)-1[x=-8]\\ y=-1\end{array}$	$(-8,-1)$
0	$\begin{array}{l}y=0·x-1\\ y=0·(0)-1[x=0]\\ y=-1\end{array}$	$(0,-1)$
4	$\begin{array}{l}y=0·x-1\\ y=0·(4)-1[x=4]\\ y=-1\end{array}$	$(4,-1)$

The points are: $(-8,-1);(0,-1);(4,-1)$

Graph of the equation: $y=-1$, this is a line parallel to x − axis.

  

Thus, the points on the graph are: $(-8,-1);(0,-1);(4,-1)$ for the equation $y=-1$