Solutions to Caribbean Algebra II Problems

Graph for the simultaneous solution problem, (4)



Graph for the velocity - time problem, (5)

Solutions to Algebra II test:

1) (a) If (3x + 1)/3 - (x - 3)/2 = 2 + (2x - 3)/3

find the value of x:

We begin by simplifying both sides and obtained lowest common denominator:

-> (6x + 2 - 3x + 9)/ 6 = (6 + 2x -3)/3 = 2(2x + 3)/6

-> 6x + 2 - 3x + 9 = 4x + 6

-> 3x + 11 = 4x + 6

-> x = 11 - 6 = 5

Which can be inserted into the original (fractional) equation to verify both sides are equal.

b) Factorize completely:

(i) 15 x² y - 20 xy²

= 5xy (3x - 4y)

2) f and g are functions defined as follows:

f: x -> 3x - 5

g: x -> x/2

a) Calculate the value of f(-3)

f(-3) = 3(-3) - 5 = -9 -5 = -14

b) Write expressions for (i) f^-1(x) and (ii) g^-1(x)

(i)the inverse function is obtained by exchanging x and y places and solving for the other variable, viz: x = 3y - 5, then y = (x + 5)/3, therefore:

f^-1(x) = (x + 5)/3

(ii) Similarly, g^-1(x) = 2x (since x = y/2 and hence y = 2x)

c) Hence or otherwise, write an expression for (gf)^-1

This is a composition function of the first two, with the order of priority in operation to the right, e.g. f ^-1 first then g^-1. Hence, combining the two:

(gf)^-1 = 2 [(x + 5)/3] = 2(x + 5)/ 3

3) This graph is easily sketched and is a downward oriented parabola with minimum at (x, y) = (-0.5, -6.2). Inspection of the graph shows that the interval of the domain (first coordinate) for which F(x) < 0 is just:    -3 <  x < 2

Over this interval the parabola defining the equation is below the zero line for y = F(x)

4) The graph for the quadratic is sketched as the parabola in the uppermost graphic.

a) The gradient for the same curve at the point where x = 2 can be found by carefully placing a ruler or straight edge at the point and bearing in the same inclination. Thereby, one estimates the ratio of the "rise" over the "run". In other words, the change in y coordinates to the change in x coordinates.  This will be found to be m = 5. (E.g. roughly 7.5/ 1.5 = 5)

b) The graph of 5y = 18x + 36 is shown as the straight line on the same axes. (To graph this, the student may recast the eqn. as: y = (18/5) x + 36/5.

c) The solutions to the simultaneous equations can be roughly estimated at the intersections of the two graphs. These will then yield the approximate x values to solve both - again very roughly, since eyeballing the values is not a "piece of cake". But the solutions are (approx.) x = -7/6 and x = 4.5.


5) a) The velocity -time graph is shown in the lower graphic at the top of this post.

b) i) The velocity after 2 seconds is read off the graph as v = 12.5 m/s (the acceleration portion of the graph crosses midway between 10 m/s and 15 m/s on the vertical scale.)

ii) The car's acceleration in the first five seconds will the v coordinate maximum (30 m/s) divided by the horizontal or time coordinate (5 s) or:  30 m/s/ 5s = 6 meters per second per second.

iii) The distance travelled in the 15 s is the total area under the v-t graph. In this case, we will add a triangle with base of 5s and altitude 30 m/s to a rectangle with height 30 m/s and width of 10 m/s. Then:

A = 1/2 (30 m/s) (5s) + (30 m/s) (10 s) = 75 m + 300 m = 375 m

iv) The average speed for the 15 seconds is the total distance divided by the time or:

V(av) =  D/ t = 375 m/ 15s  = 25 m/s


6) a) the key clue is that p varies directly with q, not inversely, so irrespective of how the numbers display in the panels, the proportion: p ~ q is the decider.

Using the values in the table we therefore set up a first proportion to get b:

2/ 8 =  6.1/ b  or 2b = 48.8 and b = 24.4

Similarly we can write:

8/ b = a/ 1.2 or 8/ 24.4 = a /  (1.2)

Cross multiplying:  8(1.2) = 24.4 a = 9.6

and a = 9.6/ (24.4) or a = 0.4 to the nearest tenth


b) This word problem can easily be solved using two expressions with x denoting pencils and y denoting erasers. Then, we can write:


7x + 5y = 11.60

5x + 3y = 7.60

Solving this system yields: x = 0.80 and y = 1.20

Or, each pencil costs 80 cents and each eraser costs $1.20. Therefore, 8 erasers costs 8 x $1.20 = $9.60.

c) The key clue is that S varies directly as (r + 1)

Then we may write:

s/ 8 =  (r + 1)/ 4

when r = 3

s/ 8 = (r + 1)/ 4

Thus, when S = 8, r + 1 = 4 and these are paired (proportionately)  in the denominators. The numerators then contain the unknowns and we will set s = 50 to find the value of r corresponding to it:

50/ 8 = (r + 1)/ 4

and: 8 (r + 1) = 4 (50) = 200

Then: 8r + 8 = 200   so: 8r = 200 - 8 = 192

r = 192/ 8  = 24


7)  This is a simple application problem. We have a floor in the shape of a rectangle with the length c meters, and the width 2 meters less.

a) i) In terms of c, the width will be:  c - 2

ii) The area will be L x W = c (c - 2)

b) If the area of the floor is  15 m²    then the equation in c would be:

c( 2- c) = 15 or:   c²  - 2c   + 15 = 0

c) The width of the floor can then be determined by factoring the quadratic above, so:

(c - 5) ( c + 3) = 0

SO: c = 5 and c = -3

We dispense with the negative root as unphysical, leaving c = 5 m

Then the width of the floor  is  (c - 2) =  5m - 2m =  3m