Given that the value of the polynomial 3Y ^ 2-2y + 5 is 3, find the value of the algebraic expression 6y ^ 2-4y + 5

Given that the value of the polynomial 3Y ^ 2-2y + 5 is 3, find the value of the algebraic expression 6y ^ 2-4y + 5

3y^2-2y+5=3
3y^2-2y=3-5=-2
6y^2-4y+5
=2(3y^2-2y)+5
=-2*2+5
=1

If the value of the algebraic expression 2Y2 + 3Y + 7 is 8, then the value of 4y2 + 6y-9 is () A. 2 B. -17 C. -7 D. 7

∵ 2Y2 + 3Y + 7 is 8,
∴2y2+3y+7=8，
∴2y2+3y=1，
∴2（2y2+3y）=2=4y2+6y，
By substituting 4y2 + 6y = 2 into 4y2 + 6y-9, we get the following results:
4y2+6y-9=2-9=-7．
Therefore, C

When x-2y = - 2, the value of the algebraic expression - 3 (2y-x) ^ 2-2x + 4Y + 8 is___

Because x-2y = - 2, so 2y-x = 2
-3(2y-x)^2-2x+4y+8
=-3(2y-x)^2-2(x-2y)+8
=-3*2^2-2*(-2)+8
=0
Happy study

If the value of the algebraic expression 4Y ^ 2-2y + 5 is 7, then the value of the algebraic expression 2Y ^ 2-y + 1 is

4y^2－2y+5=7
4y^2－2y=2
Divide both sides by 2
2y^2－y=1
2y^2－y+1=1+1=2
You don't need to find the y value, the whole substitution!
Do you understand?

If the value of algebra 2Y ^ 2 + 3Y + 7 is 8, then the value of algebraic expression 4Y ^ 2 + 6y-9 is () A.2 B.7 C.12 D.13

(2y^2+3y+7)*2-23=4y^2+6y-9
4Y ^ 2 + 6y-9 = - 7, no correct answer

It is known that the square of 2012 (x + y) and the absolute value of 1 / 2x + 3 / 2y-1 are opposite to each other. (1): find the values of X and y (2) Calculate the value of 2011 times of X + 2012 times of X

It is known that the square of 2012 (x + y) is opposite to the absolute value of 1 / 2x + 3 / 2y-1
2012(x+y)^2+|1/2x+3/2y-1|=0
x+y=0
1/2x+3/2y-1=0
The solution
x=-1
Y=1
Calculate the value of 2011 times of X + 2012 times of X
=(-1)^2011+(-1)^2012
=-1+1
=0

If the square of the absolute value + (2x + y) of X + 2y-3 is known to be 0, then x =? Y =?

The equation holds only when the absolute value and square are equal to 0
x+2y-3=0
2x+y=0
So x = - 1, y = 2

It is known that the absolute value of Y + 1 and the square of (x + 2y-1) are opposite to each other. What is the X-Power of Y?

That is, the absolute value of Y + 1 + the square of (x + 2y-1) = 0
So y + 1 = 0
x+2y-1=0
So y = - 1
X=3
(- 1 3)

It is known that the values of 2003 (x + y) square and | 1 / 2x + 3 / 2y-1 | are opposite to each other (1) calculate the value of X and Y; (2) calculate the value of x2003 power + y2004 power

(1) 2003 (x + y) ≥ 0 (x + y) ≥ 0 (x + y) ≥ 0 2003 (x + y) square and 1241 / 2x + 3 / 2y-1 | are opposite numbers, then there are 2003 (x + y) + 1241 / 2x + 3 / 2y-1 | 0, then there are 2003 (x + y) + 1241 / 2x + 3 / 2y-1 = 0, then there are 2003 (x + y) = 0 | 1 / 2x + 3 / 2y-1 = 0, so x = - 1 y = 1 (2) (2) (1) (1) (1) ^ 2003 + 1 ^ 2004 = - 1 + 1 + 1 = 0 = 0 = 0 = 0 = 1 = 0 = 1 = 1 = 0 = 1 = 1 = 0 = 1 = 1

If the absolute value of X + 5 and the square of (x + 2y-3) are opposite to each other, then x =? Y =?

The absolute value of X + 5 is opposite to the square of (x + 2y-3)
x+5=0
x+2y-3=0
x=-5
Y=4