Answer:
(a, b, c) = (-3, -2, -4)
Explanation:
You want to solve the system of 3 equations in 3 variables shown in the attached figure.
Solution
The easiest solution is one provided by a suitable calculator (attached). It tells you ...
The "work" is entering the problem into the calculator.
Ad hoc
We notice the sum of 'a' coefficients is zero, so we can add the three equations to get ...
(4a +5b -6c) +(-3a -2b +7c) +(-a +4b +2c) = (2) +(-15) +(-13)
7b +3c = -26 . . . . . simplify
Multiplying the third equation by 4 and adding that to the first equation gives ...
4(-a +4b +2c) +(4a +5b -6c) = 4(-13) +(2)
21b +2c = -50 . . . . . simplify
Now, we have two equations in two unknowns.
Multiplying the first equation above by 3 and subtracting this last equation eliminates the b term to give ...
3(7b +3c) -(21b +2c) = 3(-26) -(-50)
7c = -28 . . . . . simplify
c = -4 . . . . . . . divide by 7
Using this in the first "ad hoc" equation, we can find b:
7b +3(-4) = -26
7b = -14 . . . . . . . . add 12
b = -2 . . . . . . . . divide by 7
And the values of b and c can go into the last of the original equations to give ...
-a +4(-2) +2(-4) = -13
-a = 3 . . . . . add 16
a = -3 . . . . . multiply by -1
The solution is (a, b, c) = (-3, -2, -4).
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Additional comment
The given solution is "ad hoc" because we decide how we're going to approach the solution based on the coefficients we see. Here, we are solving by "elimination," performing as little arithmetic as possible.
The third equation suggests we could make a substitution for 'a':
a = 4b +2c +13
That substitution would give a different set of equations in 'b' and 'c' than the ones shown above.
The second attachment shows a graphical solution to the system. It shows (a, b, c) = (-3, -2, -4). The graphing program requires the use of x and y instead of 'a' and 'b' for the variables.
Four methods of solving the system have been described. Take your pick. The calculator solution was the easiest, requiring each number to be entered once.