Answer:
7; solution is not extraneous
Step-by-step explanation:
we have
[tex]\sqrt{3x+4}=5[/tex]
Solve for x
squared both sides
[tex](\sqrt{3x+4})^2=5^2[/tex]
[tex]3x+4=25[/tex]
subtract 4 both sides
[tex]3x=25-4[/tex]
[tex]3x=21[/tex]
Divide by 3 both sides
[tex]x=7[/tex]
Verify
Substitute the value of x in the original equation
[tex]\sqrt{3(7)+4}=5[/tex]
[tex]\sqrt{21+4}=5[/tex]
[tex]\sqrt{25}=5[/tex]
[tex]5=5[/tex] ----> is true
therefore
The value of x=7 is a solution, Is not a extraneous
