Answer:
[tex]\large \tt {a \: = \: 3}[/tex]
Step-by-step explanation:
- [tex]\large \tt {\sqrt{5a\: + \: 1}}[/tex] [tex]\large \tt {+2 \: = 6}[/tex]
- [tex]\large \tt {\sqrt{5a \: + \: 1}}[/tex] [tex]\large \tt {= 6 \: - 2}[/tex]
- ([tex]\large \tt {\sqrt{5a \: + \: 1}}[/tex]) [tex]\large \tt {^2 \: = 4^2}[/tex]
- [tex]\large \tt {5a \: + \: 1 = 16}[/tex]
- [tex]\large \tt {5a \: = 16 \: - \:1}[/tex]
- [tex]\large \tt {5a \: = 15}[/tex]
- [tex]\large \tt \dfrac{5a}{5}[/tex] = [tex]\large \tt \dfrac{15}{5}[/tex]
- [tex]\large \tt \purple{a \: = \:3}[/tex]
[tex]\: [/tex]
Checking:
Substitute the value of a in the equation
- [tex]\large \tt {\sqrt{5a\: + \: 1}}[/tex] [tex]\large \tt {+2 \: = 6}[/tex]
- [tex]\large \tt {\sqrt{5 \: ⋅ \: 3 \: + \: 1}}[/tex] [tex]\large \tt {+2 \: = 6}[/tex]
- [tex]\large \tt {\sqrt{15 \: + \: 1}}[/tex] [tex]\large \tt {+2 \: = 6}[/tex]
- [tex]\large \tt {\sqrt{16}}[/tex] [tex]\large \tt {+ \: 2 \: = 6}[/tex]
- [tex]\large \tt {4 \: + \: 2 \: = 6}[/tex]
- [tex]\large \tt \underline{6 \: = 6}[/tex]
Therefore, the solution or the value of the variable is acceptable.
