Question

`√(6y+7)=5+√(y-3) \\I get how to begin this but have no clue how to solve for y.`

Answer 1

y = 3 or y = 7

Explanation:

`√(6y+7)=5+√(y-3)\\\\\text{Domain:}\\\\6y+7\geq0\ \wedge\ y-3\geq0\\\\y\geq-(7)/(6)\ \wedge\ y\geq3\Rightarrow\boxed{y\geq3}`

`6y+7=6y-18+25=6(y-3)+25\\\\\text{Substitute:}\ t=y-3,\ t\geq0\\\\√(6t+25)=5+√(t)\qquad\text{square of both sides}\\\\(√(6t+25))^2=(5+√(t))^2\\\\\text{use}\ (√(a))^2=a\ \text{and}\ (a+b)^2=a^2+2ab+b^2\\\\6t+25=5^2+2(5)(√(t))+(√(t))^2\\\\6t+25=25+10√(t)+t\qquad\text{subtract 25 from both sides}`

`6t=10√(t)+t\qquad\text{subtract}\ t\ \text{from both sides}\\\\5t=10√(t)\qquad\text{divide both sides by 5}\\\\t=2√(t)\qquad\text{square of both sides}\\\\t^2=(2√(t))^2\\\\t^2=4t\qquad\text{subtract}\ 4t\ \text{from both sides}\\\\t^2-4t=0\qquad\text{distribute}\\\\t(t-4)=0\iff t=0\ \vee\ t-4=0\\\\t=0\ \vee\ t=4`

`\text{return to substitution}\\\\y-3=0\ \vee\ y-3=4\qquad\text{add 3 to both sides}\\\\y=3\ \vee\ y=7`

Answer 2


`√(6y+7) -√(y-3) =5\\(√(6y+7) -√(y-3) )^2=5^2\\√(6y+7) ^2+√(y-3) ^2-2√((6y+7)(y-3)) =25\\6y+7+y-3-25=2√((6y+7)(y-3)) \\7y-21=2√((6y+7)(y-3)) \\(7y-21)^2=(2√(6y^2-18y+7y-21) )^2\\49y^2+441-294y=4(6y^2-11y-21)\\49y^2+441-294y=24y^2-44y-84\\49y^2-24y^2+441+84-294y+44y=0\\25y^2-250y+525=0\\25(y^2-10y+21)=0\\y^2-10y+21=0\\y^2-7y-3y+21=0\\y(y-7)-3(y-7)=0\\(y-7)(y-3)=0\\y-7=0 or y-3=0\\y=7 ,y=3`

Explanation: