Question

Really struggling with these problems please help!

Solve.

`√(10y + 24) - 3 = y + 1`
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Answer 1

`y_1 = -2` and
`y_2 = 4`

Explanation:

`√(10y + 24) - 3 = y + 1`

Move the constant to the right-hand side and change their sign.

`√(10y + 24) = y + 1 + 3`

combine like terms

`√(10y + 24) - 3 = y + 4`

Square both side to remove square brackets.

10y + 24 - 3 = y²+ 8y + 16

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

10y + 24 - y² - 8y - 16 = 0

Combine like terms

10y - 8y + 24 - 16 - y² = 0

2y + 8 - y² = 0

Use commutative property to reorder the terms.

-y² + 2y + 8 = 0

Change the sign of expression.

y² -2y -8 = 0

split -2y

y² + 2y - 4y - 8 = 0

Factor out y from the first pair and -4 from the second equation.

y ( y + 2 ) - 4 ( y + 2 ) = 0

Factor out y+2 from the expression.

( y + 2 ) ( y - 4)

When the products and factors equals 0, at least one factor is 0.

y + 2 = 0

y - 4 = 0

Solve for y

y = -2

y = 4

When we plug the both solution as y we found that both is true solution of this equation.

This equation has two solutions which are -2 and 4.

Answer 2

Solution given:

`√(10y + 24) - 3 = y + 1`

keep the constant term in one side

`√(10y + 24) =y+1+3`

solve possible one

`√(10y+24)=y+4`

now

squaring both side

`(√(10y+24))²=(y+4)²`

10y+24=y²+8y+16

taking all term on one side

10y+24-y²-8y-16=0

solve like terms

8+2y-y²=0

doing middle term factorisation

8+4y-2y-y²=0

4(2+y)-y(2+y)=0

(2+y)(4-y)=0

either

y=-2

or

y=4

y=-2,4