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pashok25 [27]
3 years ago
13

Solve for y

Mathematics
2 answers:
Andreas93 [3]3 years ago
8 0

Answer:

y = 3

Step-by-step explanation:

2y - 3 = \sqrt{3y^2-10y+12}

square both sides to remove sqrt bracket

(2y - 3)^2 = ( \sqrt{3y^2-10y+12} )^2

simplify both sides

(2y - 3)(2y - 3) = 3y^2 - 10y + 12

4y^2 - 12y + 9 = 3y^2 - 10y + 12

bring all value to left side

y^2 - 2y - 3 = 0

factor

(y - 3)(y + 1)

solve for y

y = 3, y = -1

When plugged back into the equation, only y = 3 is true

Firlakuza [10]3 years ago
4 0

Answer:

y = 3

Step-by-step explanation :

2y - 3 = \sqrt{ 3y² - 10y + 12}

Swap the sides both of the equation.

\sqrt{ 3y² - 10y + 12} = 2y - 3

To remove the brackets of equations square both side and simplify .

3y² - 10y + 12 = 4y² - 12y + 9

Move the expression to left-hand side and change its sign.

3y² - 10y + 12 - 4y² + 12y - 9 = 0

collect like terms

3y² - 4y² - 10y + 12y + 12 - 9 = 0

-y² + 2y + 3 = 0

Change the sign of expression. because it helps to solve.

y² - 2y - 3 = 0

Splits the term -2y

<u>y² + y</u> <u>-</u><u>3y - 3</u> = 0

Factor out y from the first pair and -3 from second pair of expression.

y ( y + 1 ) - 3 ( y + 1) = 0

Factor out y + 1 from the expression.

( y + 1 ) ( y - 3 ). = 0

When product and factors equals 0. at least one factor is 0.

y + 1 =0

y - 3 = 0

Solve for y

y = -1 and y = 3

If we plug the 3 as y in the expression we find that y = 3 is the true solution of this expression.

This equation has one solution which is y = 3.

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