Answer :
Answer:
y = 3
Step-by-step explanation:
2y - 3 = [tex]\sqrt{3y^2-10y+12}[/tex]
square both sides to remove sqrt bracket
(2y - 3)^2 = ( [tex]\sqrt{3y^2-10y+12}[/tex] )^2
simplify both sides
(2y - 3)(2y - 3) = [tex]3y^2[/tex] - 10y + 12
[tex]4y^2[/tex] - 12y + 9 = [tex]3y^2[/tex] - 10y + 12
bring all value to left side
[tex]y^2[/tex] - 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
Answer:
y = 3
Step-by-step explanation :
[tex]2y - 3 = \sqrt{ 3y² - 10y + 12} [/tex]
Swap the sides both of the equation.
[tex]\sqrt{ 3y² - 10y + 12} = 2y - 3 [/tex]
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
y² + y -3y - 3 = 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.