Can someone correct these for me.PLZ

Problem#1
Directions solve equation
�ãx+4 = 3

My answer: x = 5

Probelm #2
Directions solve equation

�ã(4x+1) + 3 = 0

My answer x = 2

Problem #3
Directions solve equation
�ã(2y+7)+4=y
My answer y= 1 and y = 9

the symbol a is a radical it came out wrong.

the first one is correct
Use brackets to show which parts are below the square root sign

in the second:
√(4x+1)+3=0
√(4x+1)=-3
square both sides etc.

You should have been told that if you obtain algebraic answers after "squaring" , all solutions have to be verfied.
If you check your answer of x=2, it does not satisfy the original equation.
So the second equation has no real solution.

The same is true for your third equation.
Even though y=1 and y=9 are algebraic solutions, when you check the answers in the original equation, only y=9 works, the other answer is not valid.

To solve problem #2, let's go step by step:

1. Start with the equation: √(4x+1) + 3 = 0
2. Subtract 3 from both sides to isolate the square root term: √(4x+1) = -3
3. To remove the square root, square both sides of the equation: (√(4x+1))^2 = (-3)^2
4. Simplify both sides: 4x + 1 = 9
5. Subtract 1 from both sides: 4x = 8
6. Divide both sides by 4 to solve for x: x = 2

So your answer of x = 2 is algebraically correct. However, you need to verify this solution by substituting it back into the original equation. If it satisfies the equation, then it is a valid solution. In this case, when you substitute x = 2 back into the original equation, you find that it doesn't satisfy the equation. Therefore, the second equation has no real solutions.

For problem #3, the process is similar:
1. Start with the equation: √(2y+7) + 4 = y
2. Subtract 4 from both sides: √(2y+7) = y - 4
3. Square both sides to remove the square root: (√(2y+7))^2 = (y - 4)^2
4. Simplify both sides: 2y + 7 = y^2 - 8y + 16
5. Rearrange the equation to bring all terms to one side: y^2 - 10y + 9 = 0
6. Solve the quadratic equation by factoring or using the quadratic formula to find the values of y. In this case, the equation can be factored as (y - 1)(y - 9) = 0, so the solutions are y = 1 and y = 9.

Again, to verify these solutions, substitute them back into the original equation. When you do so, you'll find that only y = 9 satisfies the equation, while y = 1 does not.

Remember, it's important to check your solutions, especially when dealing with radical equations or equations involving squared terms, as extraneous solutions may arise.