Question

6j-k=20
3k+6j=11
Solve from elimination method

Answer 1

That's some nice handwriting :)

Eliminate
`k` and solve for
`j`.

`3 (6j - k) + (3k + 6j) = 3\cdot23 + 11 \\\\ \implies 18j - 3k + 3k + 6j = 69 + 11 \\\\ \implies 24j = 80 \\\\ \implies \boxed{j = (80)/(24) = \frac{10}3}`

Now solve for
`k`.

`6j - k = 23 \implies k = 6j - 23 \\\\ \implies k = 6\cdot\frac{10}3 - 23 \\\\ \implies \boxed{k = -3}`