Question

4by9+5by3+-4by5+7by9+-2by3+9by5

Answer 1

$\frac{4}{9} + \frac{5}{3} + \frac{( - 4)}{5} + \frac{7}{9} + \frac{( - 2)}{3} + \frac{9}{5}$

LCM of 3, 5 and 9 = 45

$= \frac{(4 \times 5) + (5 \times 15) \times ( - 4 \times 9) + (7 \times 5) + ( - 2 \times 15) + (9 \times 9)}{45}$

$= \frac{20 + 75 + ( - 36) + 35 + ( - 30) + 81}{45}$

$= \frac{211 + ( - 66)}{45} = \frac{211 - 66}{45}$

$= \frac{145}{45} = \frac{29}{9}$

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Answer 2

$\sf{\underline{\underline{\huge{\mathbb{SOLUTION}}}}}$

:$\sf\implies\dfrac{4}{9} + \dfrac{5}{3} + \dfrac{-4}{5} + \dfrac{7}{9} +\dfrac{-2}{3} + \dfrac{9}{5}$

:$\sf\implies \dfrac{(4 \times 5) + (5 \times 15) \times ( - 4 \times 9) + (7 \times 5) + ( - 2 \times 15) + (9 \times 9)}{45}$

:$\sf\implies \dfrac{20 + 75 + ( - 36) + 35 + ( - 30) + 81}{45}$

:$\sf\implies\dfrac{211 + ( - 66)}{45}$

:$\sf\implies \dfrac{211 - 66}{45}$

:$\sf\implies \dfrac{145}{45}$

:$\sf\implies \dfrac{29}{9}$