Solving Equations..!

Please help me to solve these questions....please explain the steps, so i could learn from them !

1)5(x + 2) + 2 = −2(x − 3) − 1

2)4/5y + 1/4(y-5) = (y +1)/6

3)x^5 − 625x = 0

4)x^4 − x^3 − 56x^2 = 0

1) First multiply the 5(x+2):
5x+10 +2 = -2(x-3) -1

Then do the same with -2(x-3):
5x+10+2= -2x +6 -1

Then do the addition and subtraction;
5x+12= -2x +5

Add 2x to both sides:
7x +12 = 5

Subtract 12

7x= -7

Divide by 7

x= -1

2 will have the same steps but you have to multiply both sides by 6 to get rid of the (y+1)/6


3 factor out a (x)
x(x^4 - 625)

Then you have a difference of perfect squares (x^4 - 625) is the same as (x^2 +25)(x^2 -25):

x(x^2+25)(x^2-25)=0

Last set each factor to 0 and find the answers

x=0 x^2+25=0 x^2-25=0

x^2= -25 x^2= 25

non real x= +/- 5 (square root)

so the answers to 3 are ( -5 , 0 , 5)

for 4, do the same but start by factoring out x^2:

x^2(x^2 -x - 56) the factor:

x^2(x-8)(x+7)

Then set each to 0

x^2=0 x-8=0 x+7=0


When factoring always check your answers when getting multiple answers. Most of the time they all work, but sometimes one doesn't work with the equation.

3)
X(X^4 - 5^4)=0 Do you remember the rule for the difference of the squares?
X(X^2 +5^2)(X^2 - 5^2)=0
X(X^2 + 5^2)(X + 5)(X - 5) = 0
This has many solutions:
X=0, X=5, X=-5, and X=5i (at least that's what I think it is)

4)
X^2(X^2 - X - 56) = 0
Now, you know the quadratic equation, right? Put the part in the brackets into the quadratic equation.

The first two are very simple.

1) 2) open the brackets and simplify.
3) Use (x²-y²) = (x+y)(x-y)
4) take x² common.