– A PHONE COMPANY CHARGES

20 CENTS FOR DAYTIME MINUTES AND FIVE CENTS

FOR NIGHTTIME MINUTES. HOW MANY DAY AND NIGHT MINUTES

WOULD YOU HAVE TO USE TO BE CHARGED MORE THAN

10 DOLLARS IN 24 HOURS? SO THIS SITUATION REPRESENTS

AN INEQUALITY IN TWO VARIABLES. SO LET’S LET D EQUAL THE NUMBER

OF DAY MINUTES, AND LET’S LET N EQUAL

THE NUMBER OF NIGHT MINUTES. AND THERE’S ONE MORE ISSUE

TO DEAL WITH. THE CHARGE PER MINUTE HERE

IS IN CENTS, BUT THEN WE WANT TO ANSWER

THE QUESTION BASED UPON BEING CHARGED

MORE THAN 10 DOLLARS. SO WE HAVE TO EITHER CONVERT

THE CENTS TO DOLLARS OR DOLLARS TO CENTS. AND TO AVOID DECIMALS, LET’S

CONVERT 10 DOLLARS TO CENTS. SINCE ONE DOLLAR=100 CENTS, THEN 10 DOLLARS WOULD BE EQUAL

TO 10 x 100 OR 1,000 CENTS. NOW, LET’S SEE IF WE CAN WRITE

AN INEQUALITY TO REPRESENT THIS SITUATION. EVERY DAYTIME MINUTE COSTS

20 CENTS. SO THE AMOUNT OF MONEY SPENT

ON DAYTIME MINUTES WOULD BE 20 CENTS X D + THE AMOUNT OF MONEY SPENT

ON NIGHTTIME MINUTES WOULD BE FIVE CENTS x THE NUMBER

OF NIGHT MINUTES OR 5 x N. WE WANT TO KNOW WHEN THIS WOULD

BE MORE THAN 10 DOLLARS IN 24 HOURS OR>1,000 CENTS. SO NOW, WE’LL GO AHEAD

AND GRAPH THIS INEQUALITY WHERE THE GRAPH WOULD SHOW

ALL THE POSSIBLE SOLUTIONS TO THIS GIVEN INEQUALITY. SO WE’LL LET THE HORIZONTAL AXIS

EQUAL THE D AXIS, WHICH REPRESENTS THE NUMBER

OF DAY MINUTES. AND WE’LL LET THE VERTICAL AXIS

BE THE N AXIS, WHICH REPRESENTS THE NUMBER

OF NIGHT MINUTES. SO WE NEED TO GRAPH

THIS INEQUALITY. SO THE FIRST STEP IS TO GRAPH

THE BORDERLINE. SO WE’LL GO AHEAD AND GRAPH

THE LINE 20D + 5N=1,000, AND BECAUSE THE ORIGINAL

INEQUALITY IS>, THE LINE WILL NOT BE PART

OF THE SOLUTION, AND TO SHOW THIS,

WE’LL MAKE A DASHED LINE. AND SINCE THIS LINE

IS IN STANDARD FORM, LET’S GO AHEAD AND FIND THE X

AND Y INTERCEPTS TO GRAPH THE LINE OR IN THIS CASE,

THE D AND N INTERCEPT. SO TO FIND THE D INTERCEPT,

WE’LL SET N=0, AND THEN TO FIND

THE N INTERCEPT, WE’LL SET D=0. SO IF WE SET N=0,

5N WOULD BE 0, AND WE’D HAVE 20D=1,000. DIVIDE BOTH SIDES BY 20,

AND WE HAVE D=50. SO THE D INTERCEPT=50,

AND NOW WE’LL SET D=0. AND IF D=0, 20D WOULD BE 0,

AND WE’D HAVE 5N=1,000. DIVIDE BOTH SIDES BY 5,

AND WE’D HAVE N=200. SO NOW WE CAN PLOT THESE

TWO POINTS AND SKETCH OUR LINE, BUT BEFORE WE DO THIS, LET’S GO

AHEAD AND SCALE OUR AXES. WE CALL THIS 200,

AND THIS WOULD BE 100. IF WE LET THIS EQUAL 50,

THEN THIS WOULD BE 25. SO NOW,

WE’LL PLOT THE POINT (50, 0), WHICH IS THE D INTERCEPT,

THIS POINT HERE. AND NOW, WE’LL PLOT THE POINT

(0, 200), WHICH IS THE N INTERCEPT,

WHICH IS THIS POINT HERE. AND NOW, WE’LL SKETCH A LINE

THROUGH THOSE TWO POINTS, AND AGAIN, WE SAID

IT WAS GOING TO BE DASHED BECAUSE OF THE VISUAL INEQUALITY

SYMBOL. SO IT’S GOING TO LOOK SOMETHING

LIKE THIS, AND NOW, WE NEED TO FIGURE OUT WHETHER WE’RE GOING TO

SHADE BELOW THE LINE OR ABOVE THE LINE. AND NOTICE HOW WE’RE ONLY

CONSIDERING THE FIRST QUADRANT, BECAUSE WE KNOW THAT

THE NUMBER OF MINUTES CAN NEVER BE NEGATIVE. NOW THAT WE HAVE THE GRAPH

OF OUR BORDERLINE, WE HAVE TO FIGURE OUT

WHETHER WE SHOULD SHADE ABOVE OR BELOW THE LINE. SO IF WE SELECT THIS POINT HERE

AS OUR TEST POINT WITH THE COORDINATES (0, 0), IT’S GOING TO BE VERY EASY TO

SEE THAT WHEN D=0 AND N=0, 0 IS NOT>1,000, AND THEREFORE, THIS POINT DOES

NOT SATISFY THE INEQUALITY. SO WE HAVE TO SHADE

THE OTHER SIDE OF THE LINE OR THIS SIDE HERE. SO ANY POINT IN THE SHADED

REGION REPRESENTS A COMBINATION OF THE NUMBER OF DAY AND NIGHT

MINUTES THAT WOULD HAVE A CHARGE

OF MORE THAN 10 DOLLARS. SO WE COULD PROBABLY FIGURE OUT

LOGICALLY THAT IF THIS WAS THE LINE WHERE

THE CHARGES WOULD BE EXACTLY 10 DOLLARS OR 1,000 CENTS, THEN ANY POINT ABOVE THIS LINE WOULD REPRESENT THE COMBINATION

OF DAY AND NIGHT MINUTES THAT WOULD BE MORE THAN

10 DOLLARS, AND THEREFORE,

WOULD BE IN THE REGION THAT WOULD REPRESENT

THE SOLUTION TO THIS LINEAR INEQUALITY. I HOPE THIS EXAMPLE HELPS.

thank you