We can solve linear systems with three variables using the following two methods: elimination and substitution.
Let us use an example to explain how each of these methods can be used to solve such linear systems.
Solve for a, b, and c in the following linear system of equations:
`bc=25+3a`
`6a+c=10+5b`
`2a+b+3c=14`
First, we will simplify the equations and rearrange them so that the first, second, and third terms contain the variables a, b, and c, respectively.
The first equation can be rewritten as follows:
`bc3a = 25` . We can further simplify it by multiplying both sides of the equation with 1:
`1(bc3a) = 1(25)` to obtain `3a+b+c = 25` .
The second equation can be rewritten as follows:
`6a5b+c = 10` .
Finally, the third equation can be written as `2a+b+3c = 14` .
Therefore, we want to solve the following system of simplified equations:
`3a+b+c = 25 (i)`
`6a5b+c = 10 (ii)`
`2a+b+3c = 14 (iii)` .
Using substitution
From equation (i), we find that `c = 253ab` . We then substitute for this value of c in equations (ii) and (iii), so that we obtain two equations in two unknowns, which we can easily solve using either elimination and substitution (in this case, we use substitution).
Thus, equations (ii) and (iii) become
`6a5b+(253ab) = 10 (ii)`
`2a+b+3(253ab) = 14 (iii)`
We simplify equation (ii) as follows:
`6a3a5bb+25=10 ` > `3a6b=1025` > `\frac{3a6b=15}{3}` (divide both sides of the equation by 3) > `a2b=5 (ii)`
Next, we simplify equation (iii) as follows:
`2a9a+b3b=1475` > `7a2b=61 (iii)` .
Therefore, we want to solve equations (ii) and (iii) using substitution. From equation (ii), we find that `a = 5+2b` . We take this value of a and substitute it into equation (iii) to obtain `7(5+2b)  2b = 61 ` > `3514b2b=61` > `3516b=61` > `16b = 6135=96` ; therefore, `16b = 96` and `b = \frac{96}{16} = 6` .
Since we'd found that `a = 5+2b` , and we now know that `b = 6` , then `a = 5 + 2(6) = 5+12 = 7` .
Since we'd found that `c=253ab` , and we now know that `a=7` and `b=6` , then `c=253(7)6=25216=2` .
Thus, using substitution, `a=7, b=6` , and `c=2` .
Using Elimination
In this method, we determine a variable to eliminate in a set of two equations and proceed to eliminate this variable to obtain a new equation, which we call equation (iv). Next, we eliminate the same variable in a different set of two equations to obtain a new equation, which we call equation (v). Finally, we solve equations (iv) and (v) to obtain solutions for the two variables they contain, and we use substitution to obtain the solution to the third variable.
Therefore, given the simplified system of equations above, we choose to eliminate the variable c in equations (i) and (ii). We do this by subtracting equation (ii) from equation (i) as follows:
`3a+b+c=25 (i)`

`6a5b+c=10 (ii)`

`3a+6b=15 (iv)`
Next, we eliminate variable c from equations (ii) and (iii) as follows:
`3(6a5b+c) = 3(10) (ii)`

`2a+b+3c=14 (iii)`

`16a16b=16 (v)`
Equation (v) can be further simplified by dividing it throughout with 16 to obtain `ab=1` .
Next, we solve equations (iv) and (v) by eliminating b as follows:
`3a+6b=15 (iv)`
+
`6(ab) = 6(1) (v)`

`3a=21` > `a = \frac{21}{3}` > `a=7`
Using equation (iv), if `a=7` , then `3(7) + 6b = 15` > `21+6b=15` and `6b=15+21=36` > `b= \frac{36}{6} = 6` .
From equation (i), `3a+b+c=25` ; however, `a=7` and `b=6` . Therefore, `c=253ab=253(7)6=2` .
Thus, using elimination, we find that `a=7, b=6` , and `c=2` .
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