Question

question 7 of 10, step 1 of 1
solve the following radical equation.
\\(\sqrt{6y + 10} + 3 = y + 2\\)
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\\(y =\\) ,

Explanation:

Step1: Isolate the radical term

Subtract 3 from both sides to isolate the square root.
$\sqrt{6y+10} = y + 2 - 3$
$\sqrt{6y+10} = y - 1$

Step2: Square both sides

Eliminate the square root by squaring each side.
$(\sqrt{6y+10})^2 = (y - 1)^2$
$6y + 10 = y^2 - 2y + 1$

Step3: Rearrange to quadratic form

Move all terms to one side to set equation to 0.
$y^2 - 2y + 1 - 6y - 10 = 0$
$y^2 - 8y - 9 = 0$

Step4: Factor the quadratic

Factor the quadratic expression.
$(y - 9)(y + 1) = 0$

Step5: Solve for y

Set each factor equal to 0 and solve.
$y - 9 = 0 \implies y = 9$
$y + 1 = 0 \implies y = -1$

Step6: Verify solutions

Check solutions in the original equation.
For $y=9$: $\sqrt{6(9)+10} + 3 = \sqrt{64} + 3 = 8 + 3 = 11$, and $9 + 2 = 11$. Valid.
For $y=-1$: $\sqrt{6(-1)+10} + 3 = \sqrt{4} + 3 = 2 + 3 = 5$, and $-1 + 2 = 1$. $5
eq 1$, so invalid.

Answer:

$y = 9$