Chapter 6.1, Problem 8P

Solve each system by graphing. Check your solution.

$y=\frac{4}{3}x-3$

$y=-\frac{2}{3}x+3$

To determine

To Graph: Solve the system of equations to find value of x and y using graphing method and check the solution.

Answer to Problem 8P

The value of $x=3$ and $y=1$ .

Explanation of Solution

Given information: $y=\frac{4}{3}x-3$ --------(1)

and $y=-\frac{2}{3}x+3$ -------(2)

Graph:

Graph the system of equations using slope and y-intercept method.

For equation (1)

  $y=\frac{4}{3}x-3$

Slope $=\frac{4}{3}$ and y intercept $=-3$

To graph the line, firstly represent the y intercept (0,-3) and using slope = $=\frac{rise}{run}$ find the rise and run.

  $Rise=4$

  $Run=3$

From y intercept (0,-3) move up 4 units and right 3 units to get another point.

Then join the points and extend the line on both ends.

For equation (2)

  $y=-\frac{2}{3}x+3$

Slope $=-\frac{2}{3}$ and y intercept $=3$

Now, represent the y intercept (0,3) on the xy plane and find another point using slope using formula $\frac{Rise}{Run}$ .

  $\begin{array}{l}Rise=-2\\ Run=3\end{array}$

From y intercept move down 2 units and right 3 units to get next point. Then join the points and extend the line on both ends.

The graph of the given linear equations are

  

Interpretation: The red line represents the graph of equation (1) and the Blue line represents graph of equation (2). The lines intersect at (3,1). So, the solution to the system of equations is (3,1).

Check:

Plug in x as 3 and y as 1 into each equation to check whether the equation is true or not.

For equation (1)

  $1=\frac{4}{3}(3)-3$

Simplify the right side of the equation

  $1=4-3$

  $1=1$

Yes! The equation is true. So, the (3,1) is solution to the equation (1)

Now for equation (2)

  $1=-\frac{2}{3}(3)+3$

  $1=-2+3$

  $1=1$

Yes! The equation is true for the solution set (3,1). So this is solution to the equation (2)

So, the (3,1) is solution to the both equations.