Chapter 3.2, Problem 63PPS

a.

To determine

To make: a table of x -values and y -values for each of the given equation.

Explanation of Solution

Given:

The given equation is:

  $\begin{array}{l}3y+x=16\\ y-2x=-4\\ y+5x=10\end{array}$

Calculation:

As the first equation is $3y+x=16$ .

Therefore, $\begin{array}{l}3y=16-x\\ \Rightarrow y=\frac{16-x}{3}\end{array}$

Now create the table so that to find the values of the equation,

$x$	$y=\frac{16-x}{3}$	$\left(x,y\right)$
$-2$	$y=\frac{16-\left(-2\right)}{3}=6$	$\left(-2,6\right)$
$-1$	$y=\frac{16-\left(-1\right)}{3}=\frac{17}{3}$	$\left(-1,\frac{17}{3}\right)$
$0$	$y=\frac{16-0}{3}=\frac{16}{3}$	$\left(0,\frac{16}{3}\right)$
$1$	$y=\frac{16-1}{3}=\frac{15}{3}=5$	$\left(1,5\right)$
$2$	$y=\frac{16-2}{3}=\frac{14}{3}$	$\left(2,\frac{14}{3}\right)$

Now, create the table by using the second equation so that to find the values of the equation.

The second equation is $y-2x=-4$ .

Therefore, $y=2x-4$

$x$	$y=2x-4$	$\left(x,y\right)$
$-2$	$y=2\left(-2\right)-4=-8$	$\left(-2,-8\right)$
$-1$	$y=2\left(-1\right)-4=-6$	$\left(-1,-6\right)$
$0$	$y=2\left(0\right)-4=-4$	$\left(0,-4\right)$
$1$	$y=2\left(1\right)-4=-2$	$\left(1,-2\right)$
$2$	$y=2\left(2\right)-4=0$	$\left(2,0\right)$

Now, create the table by using the third equation so that to find the values of the equation.

The third equation is $y+5x=10$ .

Therefore, $y=10-5x$

$x$	$y=10-5x$	$\left(x,y\right)$
$-2$	$y=10-5\left(-2\right)=20$	$\left(-2,20\right)$
$-1$	$y=10-5\left(-1\right)=15$	$\left(-1,15\right)$
$0$	$y=10-5\left(0\right)=10$	$\left(0,10\right)$
$1$	$y=10-5\left(1\right)=5$	$\left(1,5\right)$
$2$	$y=10-5\left(2\right)=0$	$\left(2,0\right)$

Hence, the above three tables gets the result of the given system of equation.

b.

To determine

To check: whether any value represents the table intersections. Also, to find the point of intersection that satisfies both the three equation.

Answer to Problem 63PPS

No point of intersection.

Explanation of Solution

Given:

The given equation is:

  $\begin{array}{l}3y+x=16\\ y-2x=-4\\ y+5x=10\end{array}$

Calculation:

Let x be the number of roller skates and y be the number of inline skates.

From part (a), it is noticed that the one point (that is $x=4$ )is common between the first and second graph, but the same is not intersect with the third graph.

Thus, $x=4$ is point of intersection for the first and second equation.

Hence, find the point of intersection that satisfies both the three equation.

c.

To determine

To graph: the three equation on a single coordinate plane.

Explanation of Solution

Given:

The given equation is:

  $\begin{array}{l}3y+x=16\\ y-2x=-4\\ y+5x=10\end{array}$

Calculation:

According to the question, draw the graph of the equation:

  

Hence, the red line represents the first equation $3y+x=16$ , blue line represents the second equation $y-2x=-4$ and green line represents the third equation $y+5x=10$ .