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Solve for a, b, c
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b=-2a-3c
Solve 2a+b+3c=0 for b.
a+3\left(-2a-3c\right)+2c=1 a+2\left(-2a-3c\right)+2c=0
Substitute -2a-3c for b in the second and third equation.
a=-\frac{7}{5}c-\frac{1}{5} c=-\frac{3}{4}a
Solve these equations for a and c respectively.
c=-\frac{3}{4}\left(-\frac{7}{5}c-\frac{1}{5}\right)
Substitute -\frac{7}{5}c-\frac{1}{5} for a in the equation c=-\frac{3}{4}a.
c=-3
Solve c=-\frac{3}{4}\left(-\frac{7}{5}c-\frac{1}{5}\right) for c.
a=-\frac{7}{5}\left(-3\right)-\frac{1}{5}
Substitute -3 for c in the equation a=-\frac{7}{5}c-\frac{1}{5}.
a=4
Calculate a from a=-\frac{7}{5}\left(-3\right)-\frac{1}{5}.
b=-2\times 4-3\left(-3\right)
Substitute 4 for a and -3 for c in the equation b=-2a-3c.
b=1
Calculate b from b=-2\times 4-3\left(-3\right).
a=4 b=1 c=-3
The system is now solved.