A) 35>4.5+1.5R where R= the number of rides purchased
7>0.9+0.3R
6.1 > 0.3R
6.1/0.3 > R
B) R < 20 1/3 rides He can go on 20 rides
Graphically, R= 20 1/3 is a vertical line. R=0 is a vertical line. Shade in all points between them, between C=35 and C=0. and make the lines solid.
C) Or a one dimensional graph, two point on the real number line, connected by a line segment. Make the points solid dots, with a line connecting them 20 1/3 to the point 0.
A) Or you much more likely want the graph of the cost C,=4.5 + 1.5R with C on the y-axis and R on the x-axis. It's a straight line from (0,4.5) to (20 1/3, 35) Shade in everything below that line and above the x-axis, between the y axis and the vertical line x=20 1/3. Or more conventionally, draw a series of lines from the the line C=4.5 +1.5R to the x axis, between the y axis and the line R=20 1/3. Every point in that quadrilateral area is a solution, a feasible point.
It's an upward sloping straight line with y intercept 4.5, the point (0,4.5) and slope = 1.5, where units are measured in dollars on each axis.
2 A = 600 -35W where W = number of weeks and A = account amount
275= 600-35W
35W = 600-275
W =(600-275)/35 = 325/35= 65/7= 9 2/7 weeks . In 9 weeks she'll
have $280 left. in 10 weeks, she'll be down to $245
Plot the amount remaining on the y axis, number of weeks W, on the x axis
It's a steeply downward sloping straight line with slope -35 and y intercept (0,600) and x intercept about (17.1,0) when her account balance reaches zero,
summer is traditionally 3 months, about June 22 to Sept. 22, or almost 13 weeks.
plot it on a real number line as a dot at W=about 13. Or connect a line from W=0 to W= about 13
