Question

solve the following equation please solve 9,10​

Answer 1

Answer:

1.(4y-5)/6-(4y+14)/3=3/2

(4y+5-8y+28)/6=3/2

4y-8y+28+5=18/2

-4y+33=9

-4y=-24

y=-24/4

y=6

2)(x+3)/2-4/3=(3x-7)/3

(3x+9-8)/6=(3x-7)/3

3(3x+1)=6(3x-7)

9x+3=18x-42

9x=45

x=45/9

x=5

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Step-by-step explanation:

Answer 2

$\huge\tt{\bold{\underline{\underline{Question᎓}}}}$

$(Q9) \frac{1}{6} (4y + 5) - \frac{2}{3} (2y + 7) = \frac{3}{2}$

$Q(10) \frac{x + 3}{2} - \frac{4}{3} = \frac{3x - 7}{3}$

$\huge\tt{\bold{\underline{\underline{Answer᎓}}}}$

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$Q(9) \frac{1}{6} (4y + 5) - \frac{2}{3} (2y + 7) = \frac{3}{2}$

$= > \frac{4y + 5}{6} - \frac{4y - 14}{3} = \frac{3}{2}$

$= > \frac{4y + 5 - 2(4y - 14)}{6} = \frac{3}{2}$

$= > \frac{4y + 5 - 8y + 28}{6} = \frac{3}{2}$

$= > \frac{ - 4y + 33}{6} = \frac{3}{2}$

$= > - 8y + 66 = 18$

$= > - 8y = 18 - 66$

$= > - 8y = - 48$

$\bold{\red{= > y = 6}}$

$Q(10) \frac{x + 3}{2} - \frac{4}{3} = \frac{3x - 7}{3}$

$= > \frac{3(x + 3) - 8}{6} = \frac{3x - 7}{3}$

$= > \frac{3x + 9 - 8}{6} = \frac{3x - 7}{3}$

$= > \frac{3x + 1}{6} = \frac{3x - 7}{3}$

$= > 3(3x + 1) = 6(3x - 7)$

$= > 9x + 3 = 18x - 42$

$= > 9x - 18x = - 42 - 3$

$= > - 9x = - 45$

$\bold{\pink{= > x = 5}}$

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