How do you evaluate $\frac{a + 3 b}{4 c}$ $(a+3b)/(4c)$ when $a = 4, b = - 2, and c = - 3$ $a = 4, b = - 2, and c = - 3$ ?

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Answer 1 · 2017-08-30T17:27:55

$\frac{1}{6}$ $1/6$

Replace each variable by the value given.

$a = 4 b = - 2 c = - 3$ $color(blue)(a = 4)" "color(red)(b = -2)" "color(forestgreen)(c =-3)$

$\frac{a + 3 b}{4 c}$ $(color(blue)(a)+3color(red)(b))/(4color(forestgreen)(c))$

$= \frac{4 + 3 (- 2)}{4 (- 3)} \leftarrow$ $=(color(blue)(4)+3color(red)((-2)))/(4color(forestgreen)((-3)))" "larr$ multiply by the brackets

$= \frac{4 - 6}{- 12}$ $=(color(blue)(4)color(red)(-6))/(color(forestgreen)(-12))$

$= \frac{- 2}{12}$ $= (-2)/12$

$= \frac{1}{6}$ $=1/6$

How do you evaluate a+3b4c(a+3b)/(4c) when a=4,b=−2,andc=−3a = 4, b = - 2, and c = - 3?