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Answer :
Given that, 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
\(\Rightarrow\)15z – 21 – 18z + 22 = 32z – 52 – 17 [Solving the bracket]
\(\Rightarrow\) -3z + 1 = 32z – 69
\(\Rightarrow\) -3z – 32z = – 69 – 1 [Transposing 32z to LHS and 1 to RHS]
\(\Rightarrow\) -35z = -70
\(\Rightarrow\) z = 2
Thus, z = 2 is the required solution.