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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(20056)   (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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