Linear equations  (Page 46/6)

We need to find the coordinates of at least two points.

We arbitrarily choose $x=-1$ , $x=0$ , and $x=1$ .

If $x=-1$ , then $y=3(-1)+2$ or $-1$ . Therefore, (–1, –1) is a point on this line.

If $x=0$ , then $y=3(0)+2$ or $y=2$ . Hence the point (0, 2).

If $x=1$ , then $y=5$ , and we get the point (1, 5). Below, the results are summarized, and the line is graphed.

Again, we need to find coordinates of at least two points.

We arbitrarily choose $x=-1$ , $x=0$ and $y=2$ .

If $x=-1$ , then $2(-1)+y=4$ which results in $y=6$ . Therefore, (–1, 6) is a point on this line.

If $x=0$ , then $2(0)+y=4$ , which results in $y=4$ . Hence the point (0, 4).

If $y=2$ , then $\mathrm{2x}+2=4$ , which yields $x=1$ , and gives the point (1, 2). The table below shows the points, and the line is graphed.

To find the x-intercept, we let $y=0$ in our equation, and solve for $x$ .

Therefore, the x-intercept is 3.

Similarly by letting $x=0$ , we obtain the y-intercept which is -2.

In higher math, equations of lines are sometimes written in parametric form. For example, $x=3+\mathrm{2t}$ , $y=1+t$ . The letter $t$ is called the parameter or the dummy variable. Parametric lines can be graphed by finding values for $x$ and $y$ by substituting numerical values for $t$ .

Let $t=0$ , 1 and 2, and then for each value of $t$ find the corresponding values for $x$ and $y$ .

The results are given in the table below.