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Solve the following systems of equations:\(\frac{4}{x}\)+ 3y = 14\(\frac{3}{x}\)+ 4y = 23 |
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Answer» \(\frac{4}{x}\)+ 3y = 14 \(\frac{3}{x}\)+ 4y = 23 Multiply eq1 by 4 and eq2 by 3 and adding ⇒ 25/x = 125 ⇒ x = 1/5 Thus, 20 + 3y = 14 ⇒ y = - 2 |
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