Question 387104
given:
(1) {{{j = 2m}}}
(2) {{{s = 3m}}}
(3) {{{s = j + 3}}}
This is 3 equations and 3 unknowns, so it's solvable
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By substituting (1) into (2):
{{{m = j/2}}}
(2) {{{s = (3/2)*j}}}
{{{j = (2/3)*s}}}
Put this into (3):
{{{s = (2/3)*s + 3}}}
{{{3s = 2s + 9}}}
{{{s = 9}}}
Steve is 9 years old
check answer:
{{{j = (2/3)*s}}}
{{{j = (2/3)*9}}}
{{{j = 6}}}
and
{{{j = 2m}}}
{{{m = j/2}}}
{{{m = 3}}}
These values check OK