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Sagot :
Answer:
[tex]y_1 = -2[/tex] and [tex]y_2 = 4[/tex]
Step-by-step explanation:
[tex] \sqrt{10y + 24} - 3 = y + 1[/tex]
Move the constant to the right-hand side and change their sign.
[tex] \sqrt{10y + 24} = y + 1 + 3 [/tex]
combine like terms
[tex] \sqrt{10y + 24} - 3 = y + 4[/tex]
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:
Solution given:
[tex] \sqrt{10y + 24} - 3 = y + 1[/tex]
keep the constant term in one side
[tex] \sqrt{10y + 24} =y+1+3[/tex]
solve possible one
[tex]\sqrt{10y+24}=y+4[/tex]
now
squaring both side
[tex](\sqrt{10y+24})²=(y+4)²[/tex]
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
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