Question 153846
Let {{{f}}} = the number of free throws he made
Let {{{t}}} = the number of three-point shots he made
Let {{{s}}} = the number of two-point shots he made
The problem says
(1) {{{f*1 + t*3 + s*2 = 385}}}
and
(2) {{{f = 3t + 11}}}
(3) {{{f = s - 26}}}
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(2) {{{f = 3t + 11}}}
Solve for {{{t}}}
{{{3t = f - 11}}}
{{{t = (1/3)*f - 11/3}}}
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(3) {{{f = s - 26}}}
Solve for {{{s}}}
{{{s = f + 26}}}
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Now plug the expressions for {{{s}}} and {{{t}}}
back into (1)
(1) {{{f*1 + t*3 + s*2 = 385}}}
{{{f*1 + ((1/3)*f - (11/3))*3 + (f + 26)*2 = 385}}}
{{{f + f - 11 + 2f + 52 = 385}}}
{{{4f = 385 + 11 - 52}}}
{{{4f = 344}}}
{{{f = 86}}}
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And since,
(2) {{{f = 3t + 11}}}
{{{86 = 3t + 11}}}
{{{3t = 75}}}
{{{t = 25}}}
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Also,
(3) {{{f = s - 26}}}
{{{86 = s - 26}}}
{{{s = 112}}}
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Kobe made {{{86}}} freethrows, {{{25}}} 3-pointers,
and {{{112}}} 2-point shots
Check the answer:
(1) {{{f*1 + t*3 + s*2 = 385}}}
(1) {{{86*1 + 25*3 + 112*2 = 385}}}
{{{86 + 75 + 224 = 385}}}
{{{385 = 385}}}
OK