how to solve this linear equation ?8k-(12k+3)=6(12k+48) ?

8k-(12k+3)=6(12k+48)

First distribute to get

8k-12k-3 = 72k+288

Now combine like terms to get

8k-12k-72k = 288+3

Now add everything to get

-76k = 291

Now divide by -76 to get

k = -291/76

Have a good day!

8k - (12k + 3) = 6(12k + 48)
8k - 12k - 3 = 72k + 288
76k = - 291
k = - 291/76

Answer: k = - 291/76 or - 3 63/76

Proof:
8(- 291/76) - (12[- 291/76] + 3) = 6(12[ - 291/76] + 48)
- 582/19 - (- 873/19 + 57/19) = 6(- 873/19 + 912/19)
- 582/19 - (- 816/19) = 6(39/19)
- 582/19 + 816/19 = 234/19
234/19 = 234/19

ok²- 8k - 40 8 = 0 you may locate out what 2 components of -40 8 upload as much as -8. it quite is 12 and -4. So, (ok+12)(ok-4)=0 sparkling up for ok by making each and each parentheses equivalent to 0. ok+12=0 | ok-4=0 ok=-12 | ok=4 ok= -12 and four

8k - (12k + 3) = 6(12k + 48)
8k - 12k - 3 = 6*12k + 6*48
-4k - 3 = 72k + 288
-4k - 72k = 3 + 288
-76k = 291
k = 291/-76
k = -291/76

8k - (12k + 3) = 6(12k + 48)
8k - 12k - 3 = 72k + 288
-4k - 3 = 72k + 288
-76k = 291
k = -291/76

- 4k - 3 = 72 k + 288
- 291 = 76 k
k = - 291 / 76

8k-(12k+3)=6(12k+48)
-4k -3 = 72k + 288
-291 = 76k

k = -291/76

Expand, put all the k terms on the left, add the coefficients algebraically, put all the constants on the right, again add algebraically, divide the sum of the constants with the sum of the coefficients of k.

Simplify the equation into:-4k-3=72k+288
76k=-291
k=-291/76