Solve for a, b and c given that:
- 2a+4b+5c= 2
- 5a+3b-2c= 13
- 3a-2b-3c= 0
One way is called the substitution method. Pick any one of the equations, say the bottom one, and solve it for one of the variables:
$\displaystyle 3a - 2b - 3c = 0$
$\displaystyle c = a - \frac{2}{3}b$
Insert this value of c into the other two equations:
$\displaystyle 2a + 4b + 5 \left ( a - \frac{2}{3}b \right ) = 2 \implies 7a + \frac{2}{3}b = 2$
$\displaystyle 5a + 3b - 2 \left ( a - \frac{2}{3}b \right ) = 13 \implies 3a + \frac{13}{3}b = 13$
So now we need to solve:
$\displaystyle 7a + \frac{2}{3}b = 2$
$\displaystyle 3a + \frac{13}{3}b = 13$
Solve one of these for one of the other unknowns, etc.
For verification I get that $\displaystyle a = 0, ~ b = 3, ~ c = -2$
-Dan