Question 597196
Hi, there--

To solve this problem, we define variables for each number. Then we create a system of three equations in three variables. Finally, we solve the system of equations to find the three numbers.
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[I] Define the variables.

Call the first number a.
Call the second number b.
Call the third number c.
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[II] Create three equations.
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The sum of the three numbers is 4. Using algebra, we write this relationship as
{{{a+b+c=4}}}
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The first number minus the second plus 5 times the third is
20. In algebra we write this relationship as
{{{a-b+5c=20}}}
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The third plus 5 times the first plus the second is -8. In algebra we write this relationship as
{{{c+5a+b=-8}}}
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Rewrite the third equation to have the variables in alphabetical order.
{{{5a+b+c=-8}}}
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[III] Solve the system of equation.
I'll use the elimination and substitution methods here. You can also use a matrix to easily solve the system. Email me at my gmail address below if you want to see the matrix method, or repost this problem and ask for a matrix method solution.
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{{{a+b+c=4}}}
{{{a-b+5c=20}}}
{{{5a+b+c=-8}}}
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Add the 1st and 2nd equations together; solve to give an equation in a and c (NOTE: b+(-b)=0.)
{{{2a+6c=24}}
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Add the 2nd and 3rd equations together to give another equation in a and c.
{{{6a+6c=12}}}
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Subtract {{{6a+6c=12}}} from {{{2a+6c=24}}}. Solve for a.
{{{-4a=12}}}
{{{a=-3}}}
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Substitute -3 for a in {{{2a+6c=24}}}. Solve for c.
{{{2(-3)+6c=24}}}
{{{(-6)+6c=24}}}
{{{6c=30}}}
{{{c=5}}}
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Substitute -3 for a and 5 for c in the first equation. Solve for b.
{{{a+b+c=4}}}
{{{(-3)+b+(5)=4}}}
{{{2+b=4}}}
{{{b=2}}}
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According to our algebra, the three numbers are -3, 2, and 5.
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[IV] Check your work.
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{{{a+b+c=4}}}
{{{(-3)+2+5=4}}}
{{{4=4}}}
Check!
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{{{a-b+5c=20}}}
{{{(-3)-2+5(5)=20}}}
{{{-5+25=20}}}
{{{20=20}}}
Bingo!
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{{{c+5a+b=-8}}}
{{{(5)+5(-3)+(2)=-8}}}
{{{5-15+2=-8}}}
{{{-8=-8}}}
Presto!
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That's it! Feel free to email me via gmail with any questions about the solution.
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Ms.Figgy
math.in.the.vortex@gmail.com