SOLUTION: 1.(5x+2)(5x-2)-(x+3)^2 my answer is (5x+2)(5x-2)=5x^2-2^2=25x-4 -(x+3)^2 = -x2+x*2*3+32= -x2+6x+9 -x2+6x+9+25x-4=-x2+31x+5. 2.solving Y=2x+1 Y=3x-1

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Question 523980: 1.(5x+2)(5x-2)-(x+3)^2
my answer is
(5x+2)(5x-2)=5x^2-2^2=25x-4
-(x+3)^2 = -x2+x*2*3+32= -x2+6x+9
-x2+6x+9+25x-4=-x2+31x+5.



2.solving Y=2x+1
Y=3x-1

when i do that with the calculater i get diffrent result then when solve it with algebraic solution
this is what i get from the calculator 1,7x;4,49y and this is what i get when i try algebraic solution 2,5x;6y
plz help and tell me what i am doing wrong.

Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
1.(5x+2)(5x-2)-(x+3)^2
my answer is
(5x+2)(5x-2)=5x^2-2^2=25x-4
-(x+3)^2 = -x2+x*2*3+32= -x2+6x+9
-x2+6x+9+25x-4=-x2+31x+5.
There are two mistakes here.  You didn't square x 
in the first step.  It should be:

(5x+2)(5x-2) = (5x)²-4 = 25x²-4


And you didn't put parentheses around the whole
expression in the next step.  It should be:

-(x+3)² = -(x²+x*2*3+3²) = -(x²+6x+9) = -x²-6x-9

-x²-6x-9+25x²-4 = 24x²-6x-13.

2.solving Y=2x+1
Y=3x-1

when i do that with the calculater i get diffrent result then when solve it with algebraic solution
this is what i get from the calculator 1,7x;4,49y and this is what i get when i try algebraic solution 2,5x;6y
plz help and tell me what i am doing wrong.
Y=2x+1  
Y=3x-1

Since the right sides both equal to Y, set them equal to each other:

2x+1=3x-1
  -x=-2
   x=2

Sustitute x=2 in the first one:

Y=2x+1
Y=2(2)+1
Y=4+1
Y=5

Solution (x,Y) = (2,5)

Edwin

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