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Quiz 3 Solutions
Problem #1 (5 points): Solve the following system: 2 x โ y = 1 3 x + y = 8.
Solution We solve the first equation for y. Since 2x โ y = 1, 2x โ 1 = y. (We added y and subtracted 1 from both sides of the equation.) Plugging the expression 2x โ 1 for y in the second equation, we compute:
3 x + y = 8 3 x + (2x โ 1) = 8 5 x โ 1 = 8 5 x = 9
x =
Since x = 95 , we can now compute y from our first calculation:
2 x โ 1 = y
2 ยท
โ 1 = y 18 5
= y 13 5
= y.
As a result, the solution to our system is
9 5 ,^
13 5
Arthur travels 30 kilometers in the same time that the Holy Grail travels 10 kilometers. If the speed of the Grail is 5 kilometers per hour slower than Arthur, find the speed of Arthur.
Solution We recall that distance is the product of speed with time. Let x be the speed of Arthur in kilometers per hour. Let t be the time (in hours) it takes Arthur to travel 30 kilometers. We are given that the speed of the grail is x โ 5 kilometers per hour. Applying the distance formula we end up with the following pair of equations:
30 = xt 10 = (x โ 5)t.
We now solve both equations for t:
30 x
= t 10 x โ 5
= t.
Thus, (^30) x = t = (^) x^10 โ 5 and so (^30) x = (^) x^10 โ 5. Taking reciprocals, we obtain:
x 30
x โ 5 10 x =
30(x โ 5) 10 x = 3(x โ 5) x = 3x โ 15 15 = 2x 15 2
= x.
Arthurโs speed is 152 kilometers per hour.
Simplify the expression (4k^3 + k^2 + k) โ (2k^3 โ 4 k^2 โ 3 k).
Solution We compute:
(4k^3 + k^2 + k) โ (2k^3 โ 4 k^2 โ 3 k) = 4k^3 + k^2 + k โ 2 k^3 + 4k^2 + 3k = 2k^3 + 5k^2 + 4k.
