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

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Answer 1 · 2016-09-25T15:35:55

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

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]}