Question

How do you graph using slope and intercept of `5x+3y=19`?

Answer 1

`y = -5/3 x + 19/3`
Plot `c " at " 6 1/3` and count `(Delta y)/(Delta x) " as " -5/3`

Explanation:

Write the equation of the straight line in the form of #y = mx + c" first

`5x +3y = 19`

`3y = -5x +19`

`y = -5/3 x + 19/3" "rarr 19/3 =6 1/3`

This means that the y-intercept is at `6 1/3`

Start the graph by marking that point.
the slope is `-5/3`

From `6 1/3`, count UP 5 units and 3 to the LEFT, mark a point.
Again, count UP 5 units and 3 to the LEFT, mark a point.

From `6 1/3`, count DOWN 5 units and 3 to the RIGHT, mark a point.
Again, count DOWN 5 units and 3 to the RIGHT, mark a point.

This will give you 5 points which you can use to draw the straight line. Remember to write the equation on the line.

graph{y= -5/3 x+ 19/3 [-11.71, 8.29, 3.1, 13.1]}