Question

Solve the equation.
k+(2-5k)(6)=k+12

Answer 1

Solve for k:
\(\large 6 (2-5 k)+k = k+12\)
Distribute \(\textrm 6\) over \(\textrm{2-5 k.}\)
\(\large 6 (2-5 k) = 12-30 k:\)
\(\large k+12-30 k = k+12\)
Combine like terms in \(\textrm{k-30 k+12.}\)
\(\large k-30 k = -29 k:\)
\(\large -29 k+12 = k+12\)
Move terms with k to the left hand side.
Subtract k from both sides:
\(\large 12+(-29 k-k) = (k-k)+12\)
Combine like terms in \(\textrm{-29 k-k.}\)
\(\large -29 k-k = -30 k:\)
\(\large -30 k+12 = (k-k)+12\)
Look for two terms that sum to zero.
\(\large k-k = 0:\)
\(\large 12-30 k = 12\)
Isolate terms with k to the left hand side.
Subtract 12 from both sides:
\(\large (12-12)-30 k = 12-12\)
Look for two terms that sum to zero.
\(\large 12-12 = 0:\)
\(\large -30 k = 12-12\)
Look for two terms that sum to zero.
\(\large 12-12 = 0:\)
\(\large -30 k = 0\)
Divide both sides by a constant to simplify the equation.
Divide both sides of \(\normalsize\textrm{ -30 k = 0 by -30:}\)
\(\large\frac{-30 k}{-30} = \frac{0}{-30}\)
Any nonzero number divided by itself is one.
\(\large\frac{-30}{-30} = 1:\)
\(\large k = \frac{0}{-30}\)
Any number times zero is zero.
\(\large\frac{0}{-30} = 0:\)

Answer 2

Which leaves you with...?

Answer 3

Thank you so much