Premium Online Home Tutors
3 Tutor System
Starting just at 265/hour
Answer :
Here we have \(\frac{x}3+1=\frac{7}{15}\)
\(\Rightarrow \frac{x}3+1-1=\frac{7}{15}-1 \quad \)(Subtracting 1 from both the sides)
\(\Rightarrow \frac{x}3=\frac{7}{15}-1 \)
\(\Rightarrow \frac{x}3=\frac{7-15}{15}\)
\(\Rightarrow \frac{x}3=\frac{-8}{15}\)
\(\Rightarrow \frac{x}3×3=\frac{-8}{15}×3\)(multiplying both the sides by 3)
\(\Rightarrow x=\frac{-8}{5}\)
So, we have \(x=\frac{-8}{5}\) as the required solution.
Check:
Keep \(x=\frac{-8}{5}\) in the given solution
L.H.S= \(\frac { x }{ 3 } + 1 =\frac{-8}{5}×\frac{1}{3}+1=\frac{-8+15}{15}=\frac{7}{15}=R.H.S\)
Hence \(x=\frac{-8}{5}\) is the required solution.