Math Help please?

Question

1.)Solve: ( x + 4 )( x - 3 ) = 8

2.)Solve: y^2 + 4y - 5 = 0

3.)Solve:( y + 5 )( y - 7 ) = -20

Answer 1

1. (x + 4)(x - 3) = 8
Let's multiply out the left side of the equation.
x^2 + 4x - 3x - 12 = 8
Combine everything on the left side and leave zero on the right side.
x^2 + x - 12 - 8 = 0
x^2 + x - 20 = 0
Now factor
(x + 5)(x - 4) = 0
Either x + 5 = 0, or x - 4 = 0
x = -5, 4

2. y^2 + 4y - 5 = 0
Let's factor out the left side
(y + 5)(y - 1) = 0
Either y+5 = 0 or y-1 = 0
y = -5, 1

3. (y + 5)(y - 7) = -20
Multiply out left side of the equation
y^2 + 5y - 7y - 35 = -20
Simplify
y^2 - 2y - 35 = -20
Add 20 to both sides
y^2 - 2y - 35 + 20 = -20 + 20
Simplify
y^2 - 2y - 15 = 0
Now factor
(y - 5)(y + 3) = 0
Either y-5=0 or y+3=0
y = 5, -3

Answer 2

1.)Solve: ( x + 4 )( x - 3 ) = 8
expand this

x^2-3x+4x-12=8

x^2+x-20=0
then break this in to two ()()

x^2+5x-4x-20=0

x(x+5)-4(x+5)=0

(x+5)(x-4)=0
then

(x+5)=0 Or (x-4)=0

x=-5 Or x=4

2.)Solve: y^2 + 4y - 5 = 0
you have to use the same method which i use in first one to solve this.i'll teach you how to do that.first multiply the nombers infrount y^2 and y^0.then by that no try to make the no you have infrount y^1.we'll do this like that.

-5*1=-5 thenmake nomber 4 using -5.so it is like this.

4=5-1 so now you can see by multiplying 5 and -1 we can get -5 and by that we can get 4 also.we can apply that to this now.

y^2+5y-y-5=0

y(y+5)-1(y+5)=0

(y+5)(y-1)=0

(y+5)=0 Or (y-1)=0

y=-5 Or y=1

3.)Solve:( y + 5 )( y - 7 ) = -20
this one also same .

y^2-7y+5y-35+20=0

y^2-2y-15=0

y^2-5y+3y-15=0

y(y-5)+3(y-5)=0

(y+3)(y-5)=0

(y+3)=0 Or (y-5)=0

y=-3 Or y=5
----------------------------------------------------------------------------

Answer 3

(x + 4)(x - 3) = 8

(x + 4)(x - 3) = X2 + 4x - 3x - 12 = x² + x - 12

x² + x - 12 = 8

x² + x - 12 - 8 = 8 - 8

x² + x - 20 = 0

(x + 5)(x - 4)

x = - 5 Or x = 4

- - - - - - - - - - - - -

y² + 4y - 5 = 0

(y + 5)(Y - 1)

y = - 5 or y = 1

- - - - - - - - - - - - -

(y + 5)(y - 7) = - 20

(y + 5)(y - 7 = y² + 5y - 7y - 35 = y² - 2y - 35

y² - 2y - 35 = - 20

y² - 2y - 35 + 20 = - 20 + 20

y² - 2y -15 = 0

(y - 5)(Y + 3)

y = 5 or y = - 3

- - - - - - - - - - -s-

Answer 4

1)
(x + 4)(x - 3) = 8
x^2 +4x -3x -12 = 8
x^2 +x -20 = 0
(x + 5)(x - 4) = 0
x = -5 or x = 4

2)
y^2 + 4y - 5 = 0
(y + 5)(y - 1) = 0
y = -5 or y = 1

3)
(y + 5)(y - 7) = -20
y^2 + 5y -7y -35 = -20
y^2 -2y -15 = 0
(y - 5)(y + 3) = 0
y = 5 or y = -3

Answer 5

1. x^2+x-20=0, x^2+5x-4x-20=0, x(x+5)-4(x+5) = 0, (x-4)(x+5) = 0, So, X=4 or x= -5.

2. Y^2+4y-5 = 0, y^2+5y-y-5=0, y(y+5)-1(y+5) = 0, (y-1)(y+5)=0, So, y=1 or y = -5.

3. y^2-2y-35 = -20, y^2 -2y -15 = 0, y^2-5y+3y-15 = 0, y(y-5)+3(y-5) = 0, (y+3)(y-5) = 0, y= -3, y = 5.

Answer 6

here is the solution for the first question step by step
1)

(x + 4)(x - 3) = 8
left hand side (LHS) = (x + 4)(x - 3)
=x^2 +x - 12 (expand brackets)

x^2 +x - 12 = 8
x^2 +x - 20=0 (get everything on one side by adding8 on both sides)
(x-4) (x+5) = 0 (factorise the quadratic)

therefore either
x-4 = 0
x=4
or
x+5 = 0
x=-5

try doign the other two ur self, its the same steps
but with the second one u dont have to expand it, its already done for u! hope this helps!

Answer 7

1)
x^2 + x - 20 = 0
(x+5)(x-4) = 0
x = 4,-5)

2) y^2 + 4y - 5 = 0
(y+5)(y-1) = 0
y = (1,-5)

3) ( y + 5 )( y - 7 ) = -20
y^2 - 2y - 15 = 0
(y-5)(y+3) = 0
y = (5,-3)

Answer 8

(x + 4) (x - 3) = 8
= (-1) x (-8)
(x + 4) (x - 3) = (-5 + 4) x ( -5 - 3)
Comparing both sides x = -5
or
(x + 4) (x - 3) = 8
= (8) x (1)
(x + 4) (x - 3) = (4 + 4) x ( 4 - 3)
Comparing both sides x = 4
Similarly u can solve others

Answer 9

2 is easy
-5

Answer 10

x=4
y=4
y=15