9y^2 + 18y - 12 = 6y
Can someone help me solve this. I am struggling. 😢
subtract 6y ... 9y^2 + 12y - 12 = 0 ... divide by 3 ... 3y^2 + 4y - 4 = 0
factor ... (3y - 2) (y + 2) = 0
to find y , set each factor equal to zero
9y^2 + 18y - 12 = 6y
9y^2 + 18y - 6y -12 = 0
9y^2 + 12y - 12 = 0
3y^2 + 4y - 4 = 0
A*C = 3(-4) = -12 = -2*6. sum = -2+6 = 4 = B.
3y^2 + (6y-2y) - 4 = 0
3y(y+2) -2(y+2) = 0
(y+2)(3y-2) = 0.
y+2 = 0, Y = -2.
3y-2 = 0, Y = 2/3.
Of course! I'm here to help. To solve this equation, we need to manipulate it to isolate the variable, which in this case is y. Let's go through the steps together:
1. Start by subtracting 6y from both sides of the equation to move all the terms to one side:
9y^2 + 18y - 12 - 6y = 0
Simplifying the left side gives us:
9y^2 + 12y - 12 = 0
2. Now, let's further simplify this quadratic equation by dividing all the terms by 3, which gives us:
3y^2 + 4y - 4 = 0
3. Next, we need to factorize the quadratic equation. Since the coefficient of y^2 is 3, the factors will have the form (3y + ___)(y + ___). We need to find what goes in the blanks.
Multiply the coefficient of y^2 by the constant term:
3 * -4 = -12
Now, we need to find two numbers that multiply to -12 and add up to the coefficient of y, which is 4. The numbers that satisfy these conditions are 6 and -2.
Therefore, we can factorize the equation as:
(3y + 6)(y - 2) = 0
4. Now, set each factor equal to zero and solve for y:
3y + 6 = 0 and y - 2 = 0
Solving these two equations gives us:
3y = -6 and y = 2
y = -2
So, the solutions to the equation are y = -2 and y = 2.
If you have any further questions, please let me know!