Q: What is the solution set for the equation y2 plus 14y equals 0?

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2y-7 = 14y+29 2y = 14y + 36 -12y = 36 -y = 3 y = -3

If: 14y = 98 Then: y = 7

The center of the circle is at (9, 7) on the Cartesian plane

6x+14y+8x-18y = 14x-4y

14y - 51 = 187 + 4ySubtract 4y from each side:10y - 51 = 187Add 51 to each side:10y = 238Divide each side by 10:y = 23.8

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It is not possible to solve a single linear equation in two variables.

2y-7 = 14y+29 2y = 14y + 36 -12y = 36 -y = 3 y = -3

x = -2 and y = 0

It equals 3 more than (14 times y)

The equation 7 x + 14 y + 10 z = 98001 has infinite solutions. For any x and y we can have a z which satisfies this equation. For example, if we choose x=1 and y=1 then z=9798 satisfies. To have a specific solution there must be three equations for three unknowns (x,y,z), which can be solved simultaneously.

If: 14y = 98 Then: y = 7

The center of the circle is at (9, 7) on the Cartesian plane

6x+14y+8x-18y = 14x-4y

If: 6+17y = 15+14y then 17y-14y = 15-6 so 3y = 9 and then y = 3

Without an equality sign and not knowing the plus or minus values of some of the given terms it can't be considered to be an equation.

Equation: x^2 + y^2 +10x -14y +10 = 0 Completing the squares: (x+5)^2 + (y-7) -25-49 +10 = 0 So: (x+5)^2 + (y-7) = 64 Therfore the circle's centre is at (-5, 7) and its radius is 8

Multiply both sides by 2y. 1/2y = 7 1 = 14y y = 1/14