How do you simplify $4 {z - 6 (r - 5) + 8 [z - 2 r + 3 (r - 1)]}$ $4{z-6(r-5)+8[z-2r+3(r-1)]}$ ?

Question

1506 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-20T19:48:30

$= 36 z + 8 r + 24$ $= 36z+8r+24$

This is all one long term, but there are 2 terms in the brackets.
Start by simplifying the innermost brackets.

$4 {z - 6 (r - 5) + 8 [z - 2 r + 3 (r - 1)]}$ $4{z-6(color(red)(r-5))+8[z-2r+3color(blue)((r-1))]}$

They are unlike terms, so use the distributive law .

$= 4 {z - 6 r + 30 + 8 [z - 2 r + 3 r - 3]} \leftarrow$ $=4{z color(red)(-6r+30)+8[z-2rcolor(blue)(+3r-3)]}" "larr$ simplify

$= 4 {z - 6 r + 30 + 8 [z + r - 3]}$ $=4{z -6r+30color(forestgreen)(+8[z+r-3])}$

$= 4 {z - 6 r + 30 + 8 z + 8 r - 24} \leftarrow$ $=4{z -6r+30color(forestgreen)(+8z+8r-24)}" "larr$ simplify

$= 4 {9 z + 2 r + 6} \leftarrow \times 4$ $=4{9z+2r+6}" "larr xx 4$

$= 36 z + 8 r + 24$ $= 36z+8r+24$

How do you simplify 4{z−6(r−5)+8[z−2r+3(r−1)]}4{z-6(r-5)+8[z-2r+3(r-1)]}?