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Answer :
Given that, 15(y – 4) – 2(y – 9) + 5(y + 6) = 0
\(\Rightarrow\)15y – 60 – 2y + 18 + 5y + 30 = 0 [Solving the brackets]
\(\Rightarrow\) 8y – 12 = 0
\(\Rightarrow\) 8y = 12 [Transposing 12 to RHS]
\(\Rightarrow\) y =\( \frac { 2 }{ 3 }\)
Hence, y = \(\frac { 2 }{ 3 }\) is the required solution.