Answer:
[tex]\sqrt{6y+7} -\sqrt{y-3} =5\\(\sqrt{6y+7} -\sqrt{y-3} )^2=5^2\\\sqrt{6y+7} ^2+\sqrt{y-3} ^2-2\sqrt{(6y+7)(y-3)} =25\\6y+7+y-3-25=2\sqrt{(6y+7)(y-3)} \\7y-21=2\sqrt{(6y+7)(y-3)} \\(7y-21)^2=(2\sqrt{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[/tex]
Step-by-step explanation:
Answer:
Step-by-step explanation:
[tex]\sqrt{6y+7}=5+\sqrt{y-3}\\\\\text{Domain:}\\\\6y+7\geq0\ \wedge\ y-3\geq0\\\\y\geq-\dfrac{7}{6}\ \wedge\ y\geq3\Rightarrow\boxed{y\geq3}[/tex]
[tex]6y+7=6y-18+25=6(y-3)+25\\\\\text{Substitute:}\ t=y-3,\ t\geq0\\\\\sqrt{6t+25}=5+\sqrt{t}\qquad\text{square of both sides}\\\\(\sqrt{6t+25})^2=(5+\sqrt{t})^2\\\\\text{use}\ (\sqrt{a})^2=a\ \text{and}\ (a+b)^2=a^2+2ab+b^2\\\\6t+25=5^2+2(5)(\sqrt{t})+(\sqrt{t})^2\\\\6t+25=25+10\sqrt{t}+t\qquad\text{subtract 25 from both sides}[/tex]
[tex]6t=10\sqrt{t}+t\qquad\text{subtract}\ t\ \text{from both sides}\\\\5t=10\sqrt{t}\qquad\text{divide both sides by 5}\\\\t=2\sqrt{t}\qquad\text{square of both sides}\\\\t^2=(2\sqrt{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[/tex]
[tex]\text{return to substitution}\\\\y-3=0\ \vee\ y-3=4\qquad\text{add 3 to both sides}\\\\y=3\ \vee\ y=7[/tex]
