Dividing x^2 - 16 by 4x - 24 gives the quotient (1/4)x + 3/2
y = (1/4)x + 3/2 would be an asymptote ( discontinuity)
Another would be a vertical discontinuity when x = 6 (this would make the denominator 4x-24 = 0.
64 questions are on the test
Q1. The answer is x = 1, y = 1, z = 0
<span>(i) -2x+2y+3z=0
</span><span>(ii) -2x-y+z=-3
</span>(iii) <span>2x+3y+3z=5
</span><span>_________
Sum up the first and the third equation:
</span>(i) -2x+2y+3z=0
(iii) 2x+3y+3z=5
_________
5y + 6z = 5
Sum up the second and the third equation:
(ii) -2x-y+z=-3
(iii) 2x+3y+3z=5
_________
2y + 4z = 2
(iv) 5y + 6z = 5
(v) 2y + 4z = 2
________
Divide the fifth equation by 2
(iv) 5y + 6z = 5
(v) y + 2z = 1
________
Multiple the second equation by -3 and sum the equation
(iv) 5y + 6z = 5
(v) -3y - 6z = -3
________
2y = 2
y = 2/2 = 1
y + 2z = 1
1 + 2z = 1
2z = 1 - 1
2z = 0
z = 0
-2x-y+z=-3
-2x - 1 + 0 = -3
-2x = -3 + 1
-2x = -2
x = -2/-2 = 1
Q2. The answer is x = -37, y = -84, z = -35
<span>(i) x-y-z=-8
(ii) -4x+4y+5z=7
(iii) 2x+2z=4
______
</span>Divide the third equation by 2 and rewrite z in the term of x:
(iii) x+z=2
z = 2 - x
______
Substitute z from the third equation and express y in the term of x:
<span>x-y-(2-x)=-8
x - y - 2 + x = 8
2x - y = 10
y = 2x - 10
______
Substitute z from the third equation and y from the first equation into the second equation:
</span><span>-4x + 4y + 5z = 7
-4x + 4(2x - 10) + 5(2 - x) = 7
-4x + 8x - 40 + 10 - 5x = 7
-x -30 = 7
-x = 30 + 7
x = -37
y = 2x - 10 = 2*(-37) - 10 = -74 - 10 = -84
z = 2 - x = 2 - 37 = -35</span>
Answer:
C
Step-by-step explanation:
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