MC1N1-Year-11-Mathematics-Extension-1-Algebra-Practice-Answer

Year 11 Mathematics Extension 1 algebra 1.1 practice-answer

1.1 practice-answer

2

Year 11 Mathematics Extension 1
Question 1 x 2
2
 x  1  3  x  1 x 1
2
x  x  1  3  x  1

x  x  1  3  x  1  0
2

3  x  1  x  x  1  0
2

 x  1 3  x  1  x   0
 x  1 2 x  3  0 algebra 3
  x  1
2
3
  x  1 , since x  1
2

Question 2

 y  3 y  1  2  y  1
2
2  y  1   y  3 y  1  0
 y  1 2  y 1   y  3
 y  1 y  5  0
2

y  1, y  5

Question 3

2 x  3x  2   0

x  0, x 

2
3

Question 4

 2 x  3 5x  3  6 5x  3
5x  3 28x  21  0

2

6  5x  3   2 x  3 5x  3  0  5x  3 30 x  18  2 x  3  0
2

3
3
  x  
4
5

Question 5
2
2
 5 x  7    3x  5 
4 x2  25x  6  0
 4 x  1 x  6  0 x  6 or x  0.25

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1.1 practice-answer

3

Year 11 Mathematics Extension 1
Question 6 y 2x  1 2  x  2  5
5
, or divide 2 x  1 by x  2

 2 x2 x2 x2 thus x  2

Question 7

2  5x  2  2
4  5x  0
4
 x0
5

Question 8

algebra

3x  7 y y

2x  8 y x
3x2  5xy  8 y 2  0
2

x
x
3   5    8  0
 y
 y x 8
  ,1 y 3

1  1  2   1  1  2   
1  1  2 1  1  2 

Question 9

2

Question 10 x40 x2  6 x  7  0 or x  4  0 x2  6 x  7  0

The second case could not hold simultaneously,
4  x  1, x  7

Question 11
2

(a)

x2 

(b)

x3 

1 
1
 x  2  92  7
2
x x 
3

(c)

1 
1
1

  x    3  x    27  9  18
3
x  x x

1
 2 1  3 1  5 1
 x  2  x  3   x  5  x 