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Question:
\(\frac{\mathrm{12} }{\mathrm{(4b-10)}} - \frac{\mathrm{8} }{\mathrm{(2b-6)}} = 0\)
\(\frac{\mathrm{12} }{\mathrm{(4b-10)}} = \frac{\mathrm{8} }{\mathrm{(2b-6)}}\)
12(2b−6) = 8(4b−10)
24b−72 = 32b−80
−72+80 = 32b−24b
8 = 8b
b = 1
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