# Trignometry

1. y=x + 3, y=x2 + 6x + 9, y= x-3

Solution: x= -3, y=0 or x=-2, y=1

y= x+3= x2 + 6x + 9

x+3= x2 + 6x + 9

0= x2+ 5x + 6

Solving by factorization

a+b=5

aXb= 6

x2+ 2x+ 3x + 6

0=x(x+2) + 3(x=2)

0=(x+3)(x+2)

x= -3, y= 0

x= -2, y= 0

• y= -x – 4, y=-x+2, y=x-3

y= -x – 4

y= -x+2

y= x-3

Solution: x=2.5 y= -0.5 or x= 0.5, y= -3.5

• 2x + -5y = 10

-5y = -2x -10

y = 0.4x + 2

Solution=  y=2, x=-5

• 2x – 4y=9

-4y= 9-2x

y=-2.25+0.5x

Solution = y= -2.25, x= 4.5

• y=3, x=4

Solution (x,y) = (4,3)

• y= √ x-4, y= √ x+3

= (y)2= (√ x-4 )2

= y2 = x-4

X= y2 +4

Y= √ x+3

(y)2= (√ x+3)2

Y2=x + 3

X=y2 – 3

• y=x2, y=x2+ 3, y= x2 – 4

y=x2

y=x2+ 3

y= x2 +- 4

• Difference between -√ (3) and √ (-3)

-√ (3)= -1.73205

√ (-3)= Impossible

• y= 3x2 – 2 between 5,5

when x= 5, y= 3(52)-2

=3(25)-2

=73

When y= 5, 5= 3x2 -2

3x2=5+2

3x2= 7

X2 =7/3

X= √7/3

1. y= √x+5, Between {-10,10}

when x =-10

y2=x+5

y2= -10 +5

y2= -5

√y2=√-5: Impossible to acquire a root of a negative number because there is no a negative square.

When y= 10

102= x +5

100= x+5

X=95

1. y=/x+5/ between {-15,5}

when x= -15

y= x+5

= -15+5

Y= -10

When y= 5

5=x-5

X= 5+5= 10

1. y=/x-5/ between {10, -5)

when x=10

y= 10-5

= 5

When y= -5

-5= x-5

X=0

1. 3x + y = 10

Y= 10-3x

Solution= when x=0, y= 10 and when y=0, x =3.3

1. (x2– 2xy – y2) (x2+2xy+y2)

Opening the brackets and putting like terms together

x2(x2+2xy+y2) = x4 +2x3y+x2y2

– 2xy (x2+2xy+y2) = -2x3y- 2x2y2 -2xy3

– y2 (x2+2xy+y2) = x2y2 +2xy3 + y4

Therefore, solution = x4 +y4

15.

1. (x- y) (x2+xy + y2)

Opening the brackets

x(x2+xy + y2) -y(x2+xy + y2)

x3 +x2y + xy2 -x2y- xy2-y3

Putting like terms together

x3 +(x2y -x2y) + (xy2 – xy2)-y3

Therefore solution= x3 -y3

1. (x+ y) (x2+xy + y2)

Opening the brackets

x(x2+xy + y2) +y(x2+xy + y2)

x3 +x2y + xy2 +x2y+xy2+y3

Putting like terms together

x3 +(x2y +x2y) + (xy2 +xy2)+y3

Therefore solution= x3 +2 x2y + 2 xy2 +y3

1. (x- y) (x2-xy + y2)

Opening the brackets

x(x2-xy + y2) -y(x2+xy + y2)

x3 -x2y + xy2 -x2y-xy2-y3

Putting like terms together

x3 -(x2y -x2y) + (xy2 -xy2)+y3

Therefore solution= x3 -2 x2y + y3

1. (x+y) (x2– xy + y2)

Opening the brackets

x(x2-xy + y2) +y(x2-xy + y2)

x3 -x2y + xy2 +x2y-xy2+y3

Putting like terms together

x3 -(x2y +x2y) + (xy2 -xy2)+y3

Therefore solution= x3 +y3

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