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Answer :
Given that, \(\frac { 9x }{ 7-6x } = 15\)
\(\Rightarrow \frac { 9x }{ 7-6x } = \frac { 15 }{ 1 }\)
\(\Rightarrow 9x = 15(7 – 6x)\quad \) [Cross-multiplication]
\(\Rightarrow 9x = 105 – 90x\quad \) [Solving the bracket]
\(\Rightarrow 9x + 90x = 105 \quad \)[Transposing 90x to LHS]
\(\Rightarrow 99x = 105\)
\(\Rightarrow x = \frac { 105 }{ 99 }\)
\(\Rightarrow x = \frac { 35 }{ 33 }\)