Q: How do you solve 3x-4y 2x-y?

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i don't get that if it is 2xy +8 or 8 + 2xy then you can't solve or simplify it any further if it were 8 x 2y then you time the 8 by the 2 to get 16y

Y=15/(2c+2x)

x2 + y2 = x2 - 2xy + y2 + 2xy = (x - y)2 + 2xy = 72 + 2*8 = 49 + 16 = 65 You could, instead, solve the two equations for x and y and substitute, but the above method is simpler.

2xy + 6 = 4xyz Divide through by 2: xy + 3 = 2xyz 3 = 2xyz - xy = y*(2xz - x) Therefore, y = 3/(2xz - x) provided that 2xz - x is not zero.

âˆš18xy = âˆš(9*2xy) = âˆš9*âˆš2xy = 3*âˆš(2xy)

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i don't get that if it is 2xy +8 or 8 + 2xy then you can't solve or simplify it any further if it were 8 x 2y then you time the 8 by the 2 to get 16y

Y=15/(2c+2x)

4x-y2=2xy 2x ? y5 if its plus its 7 xy

2xy

(2xy)x=2x2y

Since 2xy is a factor of 2x2y, it is automatically the GCF.

2xy + 6xy = 8xy

-2x - 2y = -122x + 2y = 122y = 12 - 2xy = 6 - x

x2 + y2 = x2 - 2xy + y2 + 2xy = (x - y)2 + 2xy = 72 + 2*8 = 49 + 16 = 65 You could, instead, solve the two equations for x and y and substitute, but the above method is simpler.

(x + 4y)(x - 6y)

2xy + 6 = 4xyz Divide through by 2: xy + 3 = 2xyz 3 = 2xyz - xy = y*(2xz - x) Therefore, y = 3/(2xz - x) provided that 2xz - x is not zero.

âˆš18xy = âˆš(9*2xy) = âˆš9*âˆš2xy = 3*âˆš(2xy)