What is the answer to solvefor y,sqrt(5-y)+1=x ?

The answer to solvefor y,sqrt(5-y)+1=x is y=-x^2+2x+4 {x>= 1}

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solvefor y,sqrt(5-y)+1=x

solve for

y, 5 - y + 1 = x

y = - x^{2} + 2 x + 4 {x \geq 1}

Solution steps

5 - y + 1 = x

Move

1

to the right side

5 - y = x - 1

Square both sides:

5 - y = x^{2} - 2 x + 1

5 - y = x^{2} - 2 x + 1

Solve

5 - y = x^{2} - 2 x + 1 : y = - x^{2} + 2 x + 4

y = - x^{2} + 2 x + 4

Verify Solutions:

y = - x^{2} + 2 x + 4 {x \geq 1}

The solution is

y = - x^{2} + 2 x + 4 {x \geq 1}

- 3 x + 2 x + 1 = - 1

x, 2 f = 4 x + 21

186 = 16 8^{2} + x^{2}

b - 5 = 20

x + 1 = 3 x + 2

What is the answer to solvefor y,sqrt(5-y)+1=x ?
The answer to solvefor y,sqrt(5-y)+1=x is y=-x^2+2x+4 {x>= 1}