year 11 - equations & inequalities test with ans
TRANSCRIPT
-
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
1/23
6/8/14 9:05 AM
Page 1 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
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:
Practise more questions like this
Answer 2:
Problem 3:
Solve:
Practise more questions like this
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
http://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalitieshttp://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalitieshttp://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalitieshttp://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalitieshttp://everythingmaths.co.za/http://everythingmaths.co.za/http://everythingmaths.co.za/grade-11/03-number-patternshttp://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-10.cnxmlplus -
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
2/23
6/8/14 9:05 AM
Page 2 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
Problem 4:
Solve for :
Practise more questions like this
Answer 4:
Problem 5:
Solve for :
(Show your answer correct to one decimal place.)
(Show your answer correct to two decimal places.)
Practise more questions like this
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
http://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalitieshttp://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalities -
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
3/23
6/8/14 9:05 AM
Page 3 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
Problem 6:
Solve for in terms of by completing the square:
Practise more questions like this
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
http://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalities -
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
4/23
-
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
5/23
6/8/14 9:05 AM
Page 5 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
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.
Practise more questions like this
Answer 9:
Problem 10:
For which values of will the expression be:
undefined? equal to zero?
Practise more questions like this
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
http://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalitieshttp://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalities -
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
6/23
6/8/14 9:05 AM
Page 6 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
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.
Practise more questions like this
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?
Practise more questions like this
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
http://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalitieshttp://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalities -
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
7/23
6/8/14 9:05 AM
Page 7 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
Solve for if .
Practise more questions like this
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.
Practise more questions like this
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
=
http://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalitieshttp://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalities -
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
8/23
6/8/14 9:05 AM
Page 8 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
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
-
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
9/23
6/8/14 9:05 AM
Page 9 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
Practise more questions like this
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
http://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalities -
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
10/23
6/8/14 9:05 AM
Page 10 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
Practise more questions like this
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.
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
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
http://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalities -
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
11/23
-
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
12/23
6/8/14 9:05 AM
Page 12 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
Now we substitute the values for back into the second equation to calculate the
corresponding -values
If :
If :
The solution is or . These are the coordinate
pairs for the points of intersection.
We make the subject of each equation:
Next equate the two equations and solve for :
Now we substitute the values for back into the second equation to calculate the
corresponding -values
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
-
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
13/23
6/8/14 9:05 AM
Page 13 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
If :
The solution is . These are the coordinate pairs for the
points of intersection.
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 :
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
-
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
14/23
6/8/14 9:05 AM
Page 14 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
The solution is . These are the coordinate pairs for
the points of intersection.
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 :
If :
The solution is or . These are the coordinate
pairs for the points of intersection.
We make the subject of each equation:
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
-
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
15/23
6/8/14 9:05 AM
Page 15 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
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 :
If :
The solution is or . These are the
coordinate pairs for the points of intersection.
Problem 17:
Solve the following systems of equations graphically:
Practise more questions like this
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
http://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalities -
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
16/23
6/8/14 9:05 AM
Page 16 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
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.
Practise more questions like this
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
http://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalities -
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
17/23
6/8/14 9:05 AM
Page 17 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
time. Now deduce that the total cost, , in Rands, for a 2000 km trip is given by:
Practise more questions like this
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.
Practise more questions like this
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
http://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalitieshttp://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalities -
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
18/23
6/8/14 9:05 AM
Page 18 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
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
-
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
19/23
-
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
20/23
6/8/14 9:05 AM
Page 20 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
The second value of leads to no real solution for .
Problem 23:
Solve for :
Practise more questions like this
Answer 23:
Problem 24:
Solve for in . Hence, solve for in
.
Practise more questions like this
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
http://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalitieshttp://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalities -
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
21/23
6/8/14 9:05 AM
Page 21 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
Problem 25:
Solve for :
Practise more questions like this
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.
Practise more questions like this
Answer 26:
Problem 27:
Solve for :
Practise more questions like this
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
http://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalitieshttp://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalitieshttp://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalities -
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
22/23
6/8/14 9:05 AM
Page 22 of 23http://everythingmaths.co.za/grade-11/02-equations-and-inequalities/02-equations-and-inequalities-11.cnxmlplus
Problem 28:
Solve for :
Practise more questions like this
Answer 28:
All of our textbook content (text and images) is available under the terms of the Creative Commons By- Attribution licence.
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
http://creativecommons.org/licenses/by/3.0/http://everythingmaths.co.za/@@practice/grade-11/02-equations-and-inequalitieshttp://creativecommons.org/licenses/by/3.0/http://twitter.com/siyavulahttp://facebook.com/siyavulahttp://www.psggroup.co.za/http://www.shuttleworthfoundation.org/http://www.siyavula.com/ -
8/12/2019 Year 11 - Equations & Inequalities Test With ANS
23/23
6/8/14 9:05 AM