Solving by substitution method PDF

Solving by substitution method
3x + 2y = 6
Making x the subject of the formula
3x = 6 – 2y
x = 2 - 2
3 y
Substitute to
6x – y = 9
6(2 - 2
3 y) – y = 9
12 –4y – y = 9
12 – 5y = 9
-5y = -3
Y = 3
5
Substituting to x = 2 - 2
3 y
x = 2 -
2
3∗3
5
x = 8
5
The value of the variable y
y = 3
5

d
dx [ cos (x)
1−sin( x ) ]
=
d
dx [cos ( x ) ]∗(1−sin ( x ) )−cos (x )∗d
dx [1−sin (x )]
(1−sin ( x ) )2
= (−sin ( x ) ) ( 1−sin ( x ) )−( d
dx [1 ]− d
dx [ sin ( x ) ])cos ⁡(x )
( 1−sin ( x ) )2
= cos2 ( x )−( 1−sin ( x ) ) sin ⁡( x )
( 1−sin ( x ) )2
= cos2 (x )
( 1−sin ( x ) )2 − sin ⁡( x)
1−sin ⁡( x )
= sin2 (x)−sin ( x ) +cos2 ( x)
( sin ( x ) −1 )2
X = -0.97
Substituting
= sin2 (−0.97)−sin (−0.97 ) +cos2 (−0.97)
( sin (−0.97 )−1 )2
= 0.983352931
= 0.98

Question 3
Y = 8x + 4x2
d
dx [4 x2 +8 x ]
= 4∗d
dx [x2 ]+ 8∗d
dx [ x ]
= 4*2x + 8*1
=8x +8
At x = 1
=8*1 +8
= 17
Question 4
d
dx [ x3+ 7 x+ 97]
= d
dx [ x3 ]+ 7∗d
dx [ x ]+ d
dx [97]