Question

PLEASE ANSWER FRIENDS ​

Answer 1

Answer:

$\boxed{\huge{\blue{y = 3}}}$

Step-by-step explanation:

$\bold{To \: find} = > solve \: \\ \sqrt{y + 1} \: + \: \sqrt{2y - 5} = 3$

$\bold{Concept \: used} = > \\ \boxed{\large{\green{{(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab}}} \\ \\ \: \boxed{\large{\blue{ { ( \: \sqrt{x} \: )}^{2} = x}}}$

$\bold{Solution} = > \\ \sqrt{y + 1} + \sqrt{2y - 5} = 3$

$= > \sqrt{2y - 5} = 3 - \sqrt{y + 1}$

$squaring \: both \: sides$

$= > {( \: \sqrt{2y - 5} \: )}^{2} = {( \: 3 - \sqrt{y + 1} \: ) }^{2}$

$= > 2y - 5 = {(3)}^{2} + {( \: \sqrt{y + 1} \: ) }^{2} - 2 \: (3) \: ( \sqrt{y + 1} \: )$

$= > 2y - 5 = 9 + y + 1 - 6 \sqrt{y + 1}$

$= > 2y - 5 - y - 10 = - 6 \sqrt{y + 1}$

$= > y - 15 = - 6 \sqrt{y + 1}$

$squaring \: both \: sides \: we \: get$

$= > {(y - 15)}^{2} = ( - 6 \sqrt{y + 1} \: )^{2}$

$= > {(y)}^{2} + {(15)}^{2} - 2(y) \: (15) = 36(y + 1)$

$= > {y}^{2} - 30y + 225 = 36y + 36$

$= > {y}^{2} - 30y - 36y + 225 - 36 = 0$

$= > {y}^{2} - 66y + 189 = 0$

$= > {y}^{2} - (63 + 3)y + 189 = 0$

$= > {y}^{2} - 63y - 3y + 189 = 0$

$= > y \: (y - 63) - 3 \: (y - 63) = 0$

$= > (y - 63) \: (y - 3) \: = 0$

$if \: y \: - 6 3 = 0 \\ = > y = 63 \\ if \: y - 3 = 0 \\ = > y = 3$

$now \: we \: will \: verify \: that \: \\ these \: values \: satisfy \: given \\ equation \: or \: not \:$

$for \: y \: = 63 \\ putting \: y = 63 \: in \: given \: equation \\ \sqrt{63 + 1} + \sqrt{2(63) - 5} = 3 \\ = > \sqrt{64} + \sqrt{126 - 5} = 3 \\ = > 8 + \sqrt{121} = 3 \\ = > 8 + 11 = 3 \\ = > 19 = 3 \\ which \: is \: not \: true \: so \\ y = 6 3 \: is \: not \: a \: solution \: of \: given \: equation$

$now$

$for \: y \: = 3 \\ putting \: y = 3 \: in \:given \: equation \\ \sqrt{3 + 1} + \sqrt{2(3) -5} = 3 \\ = > \sqrt{4} + \sqrt{6 - 5} = 3 \\ = > 2 + \sqrt{1} = 3 \\ = > 2 + 1 = 3 \\ = > 3 = 3 \\ which \: is \: true \: so \: y = 3 \: is \: soution \: of \: given \: equation$

Answer 2

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

Solve $\sf{\:\sqrt{(y+1)}\:+\:\sqrt{(2y-5)}\:=\:3}$

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\sf{\pink{\:\sqrt{(2y-5)}\:+\:\sqrt{(y+1)}\:=\:3}}$

$\mapsto\sf{\:\sqrt{(2y-5)}\:=\:3\:-\:\sqrt{(y+1)}}$

Squaring both side

$\mapsto\sf{\:(\sqrt{2y-5})^2\:=\:\left(3\:-\:\sqrt{(y+1)}\right)^2} \\ \\ \mapsto\sf{\:(2y-5)\:=\:(3)^2\:+\:(\sqrt{y+1})^2\:-\:2\times\:3\times (\sqrt{y+1})}$

$\mapsto\sf{\:(2y-5)\:=\:9\:+\:(y+1)\:-\:6\:\sqrt{(y+1)}} \\ \\ \mapsto\sf{\:(-y-10+2y-5)\:=\:-6\:\sqrt{(y+1)}} \\ \\ \mapsto\sf{\:(y-15)\:=\:-6\:\sqrt{(y+1)}}$

Again, squaring both side

$\mapsto\sf{\:(y-15)^2\:=\:(-6\:\sqrt{y+1})^2} \\ \\ \mapsto\sf{\:(y^2+225-30y)\:=\:36(y+1)} \\ \\ \mapsto\sf{\:(y^2-30y-36y+225-36)\:=\:0} \\ \\ \mapsto\sf{\:(y^2-66y+189)\:=\:0} \\ \\ \mapsto\sf{\:(y^2-63y-3y+189)\:=\:0} \\ \\ \mapsto\sf{\:y(y-63)-3(y-63)\:=\:0} \\ \\ \mapsto\sf{\:(y-63)(y-3)\:=\:0} \\ \\ \mapsto\sf{\:(y-63)\:=\:0\:\:or\:\:(y-3)\:=\:0} \\ \\ \mapsto\sf{\:y\:=\:63\:\:\:or\:\:y\:=\:3}$

Now, we check our solution are right or wrong ,

So, if solution is right then, all values of y satisfies above equation

If, equation is not satisfies these solution , so we can say, that our solution is wrong .

First:-

• when , y = 3

keep value of y in given equation,

$\mapsto\sf{\:(\sqrt{3+1})\:+\:(\sqrt{2\times 3 -5})\:=\:3} \\ \\ \mapsto\sf{\:(\sqrt{4})\:+\:(\sqrt{6 -5})\:=\:3} \\ \\ \mapsto\sf{\:(2)\:+\:(\sqrt{1}\:=\:3} \\ \\ \mapsto\sf{\:2+1\:=\:3} \\ \\ \mapsto\sf{\:3\:=\:3} \\ \\ \mathfrak{\bf{\:\:\:\:\:L.H.S.\:=\:R.H.S.}}$

Second:-

• when, y = 63

keep value of y in given equation,

$\mapsto\sf{\:(\sqrt{63+1})\:+\:(\sqrt{2\times 63 -5})\:=\:3} \\ \\ \mapsto\sf{\:(\sqrt{64})\:+\:(\sqrt{126 -5})\:=\:3} \\ \\ \mapsto\sf{\:(8)\:+\:(\sqrt{121})\:=\:3} \\ \\ \mapsto\sf{\:(8)+(11)\:=\:3} \\ \\ \mapsto\sf{\:19\:=\:3} \\ \\ \mathfrak{\bf{\:\:\:\:L.H.S.\:\neq\:R.H.S.}}$

Y = 63 is not satisfied given equation,

So, we can say that y = 63 is not solution of given equation,

$\Large{\underline{\underline{\mathfrak{\bf{Thus}}}}}$

• Required value of y = 3.