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Solving for 3 variables can be done in many different ways. Such a problem is commonly referred to as a "system of equations" or "simultaneous equations". A few different methods exist for solving these systems: substitution, elimination, and matrix manipulation.

Substitution: In this method each variable is written in terms of a different variable and substituted into an equation until just 1 variable remains.

[1] 2a+b+c=7 ==>a=(7-b-c)/2

[2] -2a+5b-2c=2 ==>b=(2+2c+2a)/5

[3] a+10b+c=24 ==>c=24-10b-a

Substituting [3] into [1]: a=(7-b-(24-10b-a))/2 ==> a=(9b-17)

Substituting [3] into [2]: b=(2+2(24-10b-a)+2a)/5 ==> b=(-20b+50)/5 ==> b=2

Substituting b=2 into a=(9b-17) ==> a=(9*2-17) ==> a=1

Substituting a=1 and b=2 into equation 3: 1+10*2+c=24 ==> c=3

Solution: a=1, b=2, c=3

Elimination: Add a factor of one equation to a different one to eliminate a variable

[1] 2a+b+c=7

[2] -2a+5b-2c=2

[3] a+10b+c=24

[1]+[2] ==> 6b-c=9

2*[3]+[2] ==> 25b=50 ==>b=2

Then we can substitute b=2 into 6b+c=9 to solve for c: 6*2-c=9 ==> c=3

Substituting b=2 and c=3 into [1] ==> 2a+2+3=7 ==> a=1

Matrix Manipulation (Cramer's Law): While this method seems complex, it is often easier, especially with large systems. It is also the most procedural method.

[1] 2a+b+c=7

[2] -2a+5b-2c=2

[3] a+10b+c=24

Step 1. Find the determinants* of the coefficient matrix

A=

|2 1 1|

|-2 5 -2|

|1 10 1|

[4] det(A)=2*(5+20)-1*(-2+2)+1*(-20-5)=25

Step 2. Replace the column of the variable you are trying to find in the coefficient matrix with the constants (in this case: replace 2, -2,1 with 7,2, 24) and find the determinant. We will just solve for a as an example.

Solving for a:

A'=

|7 1 1 |

|2 5 - 2 |

|24 10 1 |

[5] det(A')=7*(5+20)-1*(2+48)+1*(20-120)=25

Step 3. Divide the coefficient matrix of which a column was replaced by the coefficient matrix:

a=[5]/[4]=25/25=1

Repeat steps 2 and 3 for each variable

*There are many methods of finding a determinant, I used the expansion by minors technique

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