436 FINAL QUESTIONS ON SYSTEMS OF EQUATIONS

Solve these questions. Solutions at the end of the file.

1) State which of the following systems of lines are parallel, coincident or skew and state how many solutions each system has.

 a) . y = 2x – 5 . y = 2x + 1 b) . y = 3x + 4 . y = 2x – 4 c) . 3x + 6y = 12 . x + 2y = 4 d) . x – y + 8 = 0 2x – 10 = 2y e) 2x + y = 6 3x – y = 4

2) Solve these systems algebraically:

 a) 4x + y = 3 2x – 3y = 12 b) 3x – 5y = 35 2.5x + y = 8.5 c) . y = x 2 – 4x . y = x + 6 d) . x 2 + y 2 = 5 . y = x – 3

3)

a) Find the equation of the line through A( – 4, 3) and B(6, – 2).
b) Write the equation of the line through P(0, 2) with a slope of .
c) Find the point of intersection of the lines in parts a) and b).

4) Write the equation of the horizontal tangent line to the parabola y = x 2 – 2x – 3.

5) Viv and Jaclyn sell magazine subscriptions.
Viv's weekly salary is \$100 plus a commission of \$6.50 per subscription sold.
Jaclyn earns \$75 per week plus a \$7.00 commission for each subscription sold.

a) How many subscriptions must they sell in order to earn the same amount?
b) How much do they each earn?

6) Find the points of intersection of the circle x 2 + y 2 = 125 and the line y = – x + 15.

7) The perimeter of a rectangle is 138 cm. If the length were 3 cm. longer,
it would be twice the width. What are the dimensions of this rectangle?

8) A group of LCC students went to the greasy spoon for burgers and fries.
5 burgers and 3 fries cost \$11.25. 1 burger and 3 fries cost \$5.25.
How much would it cost for 7 burgers and 5 fries?

Solutions

1) State which of the following systems of lines are parallel, coincident or skew and state how many solutions each system has.

 a) // 0 soln b) skew 1-soln c) coincident, inf d) // 0 soln e) skew 1-soln

2) Solve these systems algebraically:

 a) 4x + y = 3 2x – 3y = 12substitution for y. . y = 3 – 4x2x – 3(3 – 4x) = 12 14x = 21 so x = 1.5 set x = 1.5, y = – 3 b) 3x – 5y = 35 2.5x + y = 8.5substitution for y. . y = 8.5 – 2.5x3x – 5(8.5 – 2.5x) = 35 15.5x = 77.5 so x = 5 set x = 5, y = – 4 c) . y = x 2 – 4x . y = x + 6comparison. .x + 6 = x² – 4x . x² – 5x – 6 = 0 (x – 6)(x + 1) = 0 if x = 6, y = 12 if x = – 1, y = 5 d) . x 2 + y 2 = 5 . y = x – 3substitution for y. . x² + (x – 3)² = 5 . x² + x² – 6x + 9 = 5 . 2x² – 6x + 4 = 0 . x² – 3x + 2 = 0 (x – 1)(x – 2) = 0 if x = 1, y = – 2 if x = 2, y = – 1

3)

 a) y = – ½ x + 1 b) y = x + 2 c) ()

4) y = x² – 2x – 3 and we know which means k = – 4
since the y-value of the vertex is – 4, the equation of the horizontal tangent is y = – 4.

5) a) Let AV = Viv's Amount and let AJ = Jaclyn's Amount, and let x = # of subscriptions sold.

AV = 100 + 6.50x ........ and ......... AJ = 75 + 7.00x
Since we want their Amount to be the same, we set AV = AJ .

100 + 6.50x = 75 + 7.00x becomes 0.50x = 25, so x = 50.

They must sell 50 subscriptions to earn the same Amount.

b) Set x = 50 in either of the 2 equations to find they earn \$425.00 each.

6) if we set y = – x + 15, then x 2 + y 2 = 125 becomes x ² + (15 – x)² = 125

We get x ² + 225 – 30x + x² = 125

So, 2x ² – 30x + 100 = 0 which is x ² – 15x + 50 = 0

or (x – 10)(x – 5) = 0 so the points of intersection are (5, 10) (10, 5)

7)

8) Let b = the cost of 1 burger ....... and ....... Let f = the cost of 1 order of fries.

5b + 3f = 11.25 .......and .......b + 3f = 5.25 which becomes b = 5.25 – 3f (substitute)

5(5.25 – 3f) + 3f = 11.25, so 26.25 – 15f + 3f = 11.25

– 12f = –15, ....so .... f = \$1.25 .......and ....... b = \$1.50

burger = \$1.50, fries = \$1.25, 7 burgers + 5 fries = \$16.75

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