Question

If (3y-9)^2+7=43, then what is(are the values of y?​

Answer 1

Answer:

y = 1 and y = 5

Step-by-step explanation:

$( 3 y -9)^2 + 7 = 43\\\\((3y)^2 + (-9)^2 + 2(3y)(-9)) = 43 - 7\\\\(9y^2 + 81 -54y) = 36\\\\9y^2 -54y + 81 -36 = 0\\\\9y^2 -54y + 45 = 0\\\\9(y^2 -6y + 5) = 0\\\\y^2 - 6y + 5 = 0\\\\y^2 - 5y - y + 5 = 0\\\\y(y - 5) -1(y - 5 ) = 0 \\\\( y - 1 )( y - 5) = 0\\\\y = 1 \ , \ y = 5$

Answer 2

The values of y in the equation (3y-9)^2+7 = 43 are 1 and 5

How to solve for y?

The equation is given as:

(3y-9)^2+7 = 43

Subtract 7 from both sides of the equation

(3y - 9)^2 = 36

Take the square root of both sides

3y - 9 = ±6

Add 9 to both sides

3y = 9 ± 6

Divide through by 3

y = 3 ± 2

Expand

y = (3 - 2 or 3 + 2)

Evaluate

y = 1 or 5

Hence, the values of y are 1 and 5

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