How do you solve $4 t + \frac{1}{3} = t + \frac{5}{6} + t - \frac{3}{6}$ $4t + 1/3 = t + 5/6 + t - 3/6$ ?

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Answer 1 · 2016-04-15T13:57:28

$t = 0$ $t = 0$

$4 t + \frac{1}{3} = t + \frac{5}{6} + t - \frac{3}{6}$ $color(blue)(4t) + 1 /3 =color(blue)(t) + 5/6 +color(blue)(t) - 3/6$

$4 t - t - t = \frac{5}{6} - \frac{3}{6} - \frac{1}{3}$ $color(blue)(4t - t -t )= 5/6 - 3/6 - 1 /3$

$4 t - 2 t = \frac{5}{6} - \frac{3}{6} - \frac{1}{3}$ $color(blue)(4t - 2t )= 5/6 - 3/6 - 1 /3$

$2 t = \frac{5}{6} - \frac{3}{6} - \frac{1}{3}$ $2t = 5/6 - 3/6 - 1 /3$

The L.C.M of the denominators $(6, 6$ $( 6 ,6$ and $3)$ $3)$ of the L.H.S $= 6$ $=6$

$2 t = \frac{5}{6} - \frac{3}{6} - \frac{1 \cdot 2}{3 \cdot 2}$ $2t = 5/6 - 3/6 - (1 *2 ) /(3 *2 )$

$2 t = \frac{5}{6} - \frac{3}{6} - \frac{2}{6}$ $2t = 5/6 - 3/6 - (2 ) /(6 )$

$2 t = \frac{5 - 3 - 2}{6}$ $2t = (5 -3 -2 ) /6$

$2 t = \frac{5 - 5}{6}$ $2t = (5 -5 ) /6$

$2 t = \frac{0}{6}$ $2t = (0) /6$

$2 t = 0$ $2t = 0$

$t = \frac{0}{2}$ $t = 0 / 2$

$t = 0$ $t = 0$

How do you solve 4t+13=t+56+t−364t + 1/3 = t + 5/6 + t - 3/6?