Year 11 - Equations & Inequalities Test With ANS

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EV ERYTHIN G MATHS

End of chapter exercises

Problem 1:

Solve: . Give your answer correct to two decimal places.

Practise more questions like this

Answer 1:

Problem 2:

Solve:


Answer 2:

Problem 3:

Solve:


Answer 3:

x 1 = 0x2

x

x = 1,62

= b 4acb2

2a

= ( 1) ( 1 4(1)( 1))2

2(1)

= 1 1 + 4

2

= 1 5

2 or x = 0,62

16(x + 1) = ( x + 1)x2

16(x + 1)16(x + 1) (x + 1)x 2

(16 )(x + 1)x 2= 16x 2

x = 4

= ( x + 1)x 2

= 0

= 0 or x = 1 or x = 1

+ 3 + = 7y 2 12

+ 3y 2
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Problem 4:

Solve for :


Answer 4:

Problem 5:

Solve for :

(Show your answer correct to one decimal place.)

(Show your answer correct to two decimal places.)


Answer 5:

Let + 3y 2

Restriction: + 3y 2Therefore k

k + 12

k+ 12k2

7k + 12k2

(k 3)( k 4)k = 3

+ 3 = 3y 2

= 0y 2 y = 0

= k 0 0

= 7

= 7 k= 0= 0 or k = 4 or + 3 = 4y 2

or = 1y 2 or y = 1

x 2 5 12 = 0x4 x2

2 5 12x 4 x 2( 4)(2 + 3)x 2 x 2

= 4x 2

x

= 0= 0

or = (no real solution)x 2 32

= 2

x

x (x 9) + 14 = 0

x = 3x2

x + 2 = 6x

+ = 11

x + 12x

x 1

x (x 9) + 14 9x + 14x 2

(x 7)(x 2)x = 7

= 0= 0= 0 or x = 2
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Problem 6:

Solve for in terms of by completing the square:


Answer 6:

xx 2 x 3x 2

x

x = 2,3

= 3= 0

= ( 1) ( 1 4(1)( 3))2

2(1)

= 1 1 + 12

2

= 1 13

2 or x = 1,3

x + 2

+ 2 x 6x 2

x

x = 1,65

= 6x

= 0

= 2 4(1)( 6)22

2(1)

= 2 4 + 24

2

= 2 28

2

= 2 2 7

2= 1 7 or x = 3,65

+1

x + 12x

x 1(x 1) + 2 x (x + 1)

x 1 + 2 + 2 xx2

+ 3 xx 2x (x + 3)

x = 0

= 1

= ( x + 1)( x 1)

= 1x2

= 0= 0 or x = 3

x p px 4 = 0x2
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Problem 9:

An equation of the form is written on the board. Saskia and Sven copy itdown incorrectly. Saskia has a mistake in the constant term and obtains the solutions ! 4and 2. Sven has a mistake in the coefficient of and obtains the solutions 1 and ! 15.Determine the correct equation that was on the board.


Answer 9:

Problem 10:

For which values of will the expression be:

undefined? equal to zero?


Answer 10:

For the expression to be undefined the denominator must be equal to 0. This meansthat and therefore

=x 1 p + 144 + p2

8x 1 x2

p + 144 + p2

8 p 144 + p2

8 p + + p +144 + p2

144 + p2

+ 144 p2

+ 144 p2

p2

p

and =x 2 p 144 + p2

8= 5

= 5

= 40

= 20= 400= 256= 16

+ bx + c = 0x2

x

Saskia:(x + 4)( x 2)

+ 2 x 8x 2 a = 1

Sven:(x 1)( x + 15)

+ 14 x 15x 2 c

Correct equation:+ 2 x 15x 2

(x + 5)( x 3) correct roots are x = 5

= 0= 0 and b = 2

= 0= 0= 15

= 0= 0 and x = 3

b 5b + 6b2

b + 2

b + 2 = 0 b = 2


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We simplify the fraction:

Therefore or will make the expression equal to 0.

Note that we cannot have as that will make the denominator 0 and the whole

expression will be undefined.

Problem 11:

Given solve for if:

is a real number. is a rational number. is an irrational number. is an integer.


Answer 11:

We first note that the restriction is: .

Next we note that for the fraction to equal 0 the numerator must equal to 0. This gives:

Now we need to decide which of these answers meet the criteria given:

All three solutions are real. There are no integer solutions.

Problem 12:

Given , for which value(s) of will the expression be:

equal to zero? defined?


Answer 12:

The expression will be equal to 0 when the numerator is equal to 0. This gives .

The expression is undefined when the denominator is equal to 0. So the expression will

be defined for all values of except where .

Problem 13:

5b + 6b2

b + 2(b 2)( b 3)

b + 2

= 0

= 0

b = 2

b = 3

b = 2

= 0( 6)(2x + 1)x2

x + 2 x

x x x x

x 2

( 6)(2x + 1)x 2

x = 1

2

= 0

or x = 6

x = 12 x = 6

(x 6)12

+ 3x2 x

x = 6

x x = 3


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Solve for if .


Answer 13:

We first note that the restriction is: .

Next we note that for the fraction to equal 0 the numerator must equal to 0. This gives:

Now we draw up a table of signs to find where the function is positive:

Table 1

Critical values

undef0

undef 0

From this we see that is the solution.

Problem 14:

Abdoul stumbled across the following formula to solve the quadratic equation

in a foreign textbook.

Use this formula to solve the equation: .

Solve the equation again, using factorisation, to see if the formula works for this

equation.

Trying to derive this formula to prove that it always works, Abdoul got stuck along the

way. His attempt is shown below:

Complete his derivation.


Answer 14:

a 08 2a a 3

a 3

8 2a 8 2a

a

= 0= 0= 4

a = 3 a = 4

a 3

+ + +a 4 +

+

a 4

a + bx + c = 0x2

x = 2c

b 4acb2

2 + x 3 = 0x2

a + bx + cx2 = 0 \8 pt ]a + +bx

cx2

= 0 Divided by where x 0\8 pt ] + + ax2 c

x2bx

=


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Problem 15:

Solve for :

x =

x = = 64

32

2c

b 4acb2

= 2( 3)

(1) 4(2)( 3)12

=

6 1 25

= 6 1 5

or x = = 1 6 6

2 + x 3x 2(2x + 3)( x 1)

x = 32

= 0= 0

or x = 1

a + bx + cx 2

a + +bx

cx 2

+ + ac

x 2bx

+ +1x 2

bcx

ac

+1x 2

bcx

+ +1x 2

bcx

b2

4c2

( +

)1

x

b

2c

2

+1x

b2c

1x

1x

x

= 0

= 0

= 0

= 0

= ac

= +ac

b2

4c2

= = 4ac + b2

4c2 4acb2

4c2

= 4acb2

4c2

=

b2c

4acb2

4c2

=

b 4acb2

2c

= 2c

b 4acb2

x

14

x 3

< 14

(x 3)2

> 32x 2x 3


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Answer 15:

or or

Problem 16:

Solve the following systems of equations algebraically. Leave your answer in surd form,

where appropriate.

< 0 3

(x 3)( x + 1)

< 4(2x 3)2

2x 15 x

x

0+ 3x23x 2

x 2 3x

0+ 3 x 4x2

5 + x4

1x 23 x

x < 1 x > 3 0,5 < x < 2,5 x 3 or 0 < x 52 x < 23 1 x < 0

x 3 4 x 1 2 x < 312

y 2x = 0

y 2x + 3 = 0x2

a 3b = 0

a + 4 = 0b2

y 5x = 0x2

10 = y 2x

p = 2 + q 3 p2

p 3q = 1

a = 0b2

a 3b + 1 = 0

a 2b + 1 = 0

a 2 12b + 4 = 0b2

y + 4 x 19 = 08y + 5 101 = 0x2


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Answer 16:

We make the subject of each equation:

Next equate the two equations and solve for :

Now we substitute the values for back into the first equation to calculate the

corresponding -values

If :

This gives the point .

If :

This gives the point .

The solution is . These are the coordinate pairs for the

points of intersection.




a + 4 18 = 0

2a + 5 57 = 0b2

y

y 2xy

= 0= 2 x

y 2x + 3x 2

y = 0= + 2 x 3x 2

x

2x0

x 2x

= + 2 x 3x 2

= 3x 2= 3= 3

xy

x = 3

y = 2 3

( ;2 )3 3

x = 3

y = 2( )3

( ; 2 )3 3

x = and y = 23 3

a

a 3ba

= 0= 3 b

a + 4b2

a= 0= 4b2

b

3b0

(b 4)( b + 1)b = 4

= 4b2= 3b 4b2= 0

or b = 1

b


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Now we substitute the values for back into the second equation to calculate the


If :

If :

The solution is or . These are the coordinate

pairs for the points of intersection.



Now we substitute the values for back into the second equation to calculate the


If :

2 + p + 3 p2

6 + 3 p + 9 p2

6 4 p 10 p2

3 2 p 5 p2(3 p 5)( p + 1)

p = 5

3

= p 1

3= p 1= 0= 0= 0

or p = 1

pq

p = 53

q = 1533

= 29

p = 1

q = ( 1) 1

3=

23

p = and q =5329 p = 1 and q =

23

a

a b2a

= 0= b2

a 3b + 1a

= 0= 3 b 1

b

b2

3b + 1b2

b

= 3 b 1= 0

= ( 3) ( 3 4(1)(1))2

2(1)

= 3 5

2

ba

b = 3+ 5 2


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If :

The solution is . These are the coordinate pairs for the

points of intersection.





If :

If :

a = 3( ) 13 + 5 2=

7 + 3 5 2

b = 3 5 2

a = 3( ) 13 5 2=

7 3 5 2

b = and a =3 5 273 5

2

a

a 2b + 1a = 0= 2 b 1

a 2 12b + 4b2

a= 0= 2 + 12 b 4b2

b

2b 12 + 10 b 5b2

b

= 2 + 12 b 4b2

= 0

=

(10) (10 4(2)( 5))2

2(2)

= 10 140

4

ba

b = 10+ 140 4

a = 2

( ) 1

10 + 140

4

= 24 + 2 140

4

= 12 + 140

2

b = 10 140 4


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The solution is . These are the coordinate pairs for

the points of intersection.





If :

If :

The solution is or . These are the coordinate

pairs for the points of intersection.


a = 2( ) 1 10 140

4

= 24 2 140

4

= 12 140

2

b = and a = 10 140 4 12 140

2

y

y + 4 x 19y

= 0= 4x + 19

8y + 5 101x 2

8y

y

= 0= 5 + 101x 2

= 5 + 101x 2

8

x

4x + 19

32x + 1525 32x + 51x 2

x

x = 3,4

= 5 + 101x 2

8= 5 + 101x 2

= 0

= ( 32) ( 32 4(5)(51))2

2(5)

= 32 1024 1020

10

= 32 4

10 or x = 3

xy

x = 3,4

y = 4(3,4) + 19= 5,4

x = 3

y = 4(3) + 19= 5

x = 3,4 and y = 5,4 x = 3 and y = 5

a


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If :

If :

The solution is or . These are the

coordinate pairs for the points of intersection.

Problem 17:

Solve the following systems of equations graphically:


Answer 17:

a + 4 b 18a

= 0= 4b + 18

2a + 5 57b22a

a

= 0= 5 + 57b2

= 5 + 57b2

2

a

4b + 18

8x + 365 8x 21b2

b

b = 3

= 5 + 57b2

2= 5 + 57b2= 0

= ( 8) ( 8 4(5)( 21))2

2(5)

= 8 64 + 420

10

= 8 484

10 or b = 1,4

ba

b = 3

a = 4(3) + 18= 6

b = 1,4

y = 4( 1,4) + 18= 23,6

b = 1,4 and a = 23,6 b = 3 and a = 6

2y + x 2 = 0

8y + 8 = 0x2

y + 3 x 6 = 0

y = + 4 4xx2


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Problem 18:

A stone is thrown vertically upwards and its height (in metres) above the ground at time (in

seconds) is given by:

Find its initial height above the ground.


Answer 18:

The initial height occurs when . Substituting this in we get:

Problem 19:

After doing some research, a transport company has determined that the rate at which

petrol is consumed by one of its large carriers, travelling at an average speed of km per

hour, is given by:

Assume that the petrol costs R 4,00 per litre and the driver earns R 18,00 per hour of travel

t

h (t ) = 35 5 + 30 tt2

t = 0

h (t)

h (0)

= 35 5 + 30 tt2

= 35 5(0 + 30(0))2= 35 m

x

P (x ) = + litres per kilometre55

2x

x

200


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time. Now deduce that the total cost, , in Rands, for a 2000 km trip is given by:


Answer 19:

Problem 20:

Solve the following quadratic equations by either factorisation, completing the square or byusing the quadratic formula:

Always try to factorise first, then use the formula if the trinomial cannot be factorised.

Solve some of the equations by completing the square.


Answer 20:

C

C (x ) = + 40 x256\ 000

x

C (x ) = 4 2000 ( + )+ 18 552x x200 2000x= + 40 x +220\ 000

x36\ 000

x

= + 40 x256\ 000

x

4 41y 45 = 0y 2

16 + 20 x = 36x2

42 +104 p + 64 = 0 p2

21y + 3 = 54 y 2

36 + 44 y + 8 = 0y 2

12 14 = 22 y y 2

16 + 0 y 81 = 0y 2

3 + 10 y 48 = 0y 2

63 5 = 26 y y 2

2 30 = 2x2

2 = 98y 2

4 41y 45y 2

4 + 41 y + 45y 2

(4y + 5)( y + 9)

y = 54

= 0= 0= 0

or y = 9


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16 + 20 x 36x 24 + 5 x 9x 2

(4x + 9)( x 1)

x = 94

= 0= 0= 0

or x = 1

42 + 104 p + 64 p2

21 + 52 p + 32 p2

(7 p + 8)(3 p + 4)

p = 87

= 0

= 0= 0

or p = 43

54 + 21 y + 3y 2

18 7y 1y 2(9y + 1)(2 y 1)

y = 19

= 0= 0= 0

or y = 12

36 + 44 y + 8y 2

9 + 11 y + 2y 2(9y + 2)( y + 1)

y = 29

= 0= 0= 0

or y = 1

12 22y 14y 2

6 11y 7y 2(3y 7)(2y + 1)

y = 73

= 0= 0= 0

or y = 12

16 + 0 y 81y 2

(4y 9)(4y + 9)

y = 94

= 0= 0

or y = 94

3 + 10 y 48y 2

(3y 8)( y + 6)

y = 83

= 0= 0

or y = 6

5 26y + 63y 2

5 + 26 y 63y 2(5y 9)( y + 7)

y = 95

= 0= 0= 0

or y = 7

2 30x 22 32x 2

16x 2(x 4)( x + 4)

x

= 2= 0= 0= 0= 4


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The second value of leads to no real solution for .

Problem 23:

Solve for :


Answer 23:

Problem 24:

Solve for in . Hence, solve for in

.


Answer 24:

We note that is a common part and so we let . Now we note that

this gives the same equation as above ( ). We have solved this and

so we can use the solution to solve for .

y y 2

y + 2y 2

y

= 2= 0

= ( 1) ( 1 4(1)(2))2

2(1)

= 1 7

2

x y

x x = + 28 x

xx 2

(x 2)2

4x + 4x 2 3x 4x 2

(x 4)( x + 1)x = 4

= + 28 x = 8 x = 8 x= 8 x= 8 x= 0 or x = 1

y 4 + 8 y 3 = 0y 2 p4( p 3 8( p 3) + 3 = 0)2

4 8y + 3y 2

(2y 3)(2y 1)

y = 3

2

= 0= 0

or y = 1

2( p 3) y = p 3

4 8y 3 = 0y 2 p

p 3

p

p

= 32

= + 332

= 92


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Problem 25:

Solve for :


Answer 25:

Problem 26:

Without solving the equation , determine the value of . Now solve

and use the result to assess the answer obtained in the question above.


Answer 26:

Problem 27:

Solve for :


Answer 27:

This has a common part of so we replace that with :

Now we can use this result to solve for :

p 3

p

p

= 12

= + 312

= 72

x 2(x + 3 = 9)12

2(x + 3)12

(x + 3)12

(x + 3)

x

= 9

= 92

= 814

= 3814

= 69

4

x + = 31

x

+x2 1

x2

x + = 31x

7

y 5(y 1 5 = 19 (y 1)2 )2

(y 1)2 k

5k 56k

k

= 19 k= 24= 4

y
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Problem 28:

Solve for :


Answer 28:

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Embedded videos, simulations and presentations are not necessarily, please check their licence. All rights are reserved for content delivered by the Intelligent Practice service.

(y 1)2

2y + 2y 2

2y 2y 2

y

= 4= 4= 0

= ( 2) ( 2 4(1)( 2))2

2(1)

= 2 12

2

t 2t (t ) = + 232

32 3tt2

t = , t = 1 or t =123 33

4
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Year 11 - Equations & Inequalities Test With ANS