1. ## Simultaneous equations

Solve for a, b and c given that:
• 2a+4b+5c= 2
• 5a+3b-2c= 13
• 3a-2b-3c= 0

2. Originally Posted by oldschool999
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:
$3a - 2b - 3c = 0$

$c = a - \frac{2}{3}b$

Insert this value of c into the other two equations:
$2a + 4b + 5 \left ( a - \frac{2}{3}b \right ) = 2 \implies 7a + \frac{2}{3}b = 2$

$5a + 3b - 2 \left ( a - \frac{2}{3}b \right ) = 13 \implies 3a + \frac{13}{3}b = 13$

So now we need to solve:
$7a + \frac{2}{3}b = 2$

$3a + \frac{13}{3}b = 13$

Solve one of these for one of the other unknowns, etc.

For verification I get that $a = 0, ~ b = 3, ~ c = -2$

-Dan

3. Originally Posted by oldschool999
Solve for a, b and c given that:
• 2a+4b+5c= 2
• 5a+3b-2c= 13
• 3a-2b-3c= 0
What methods have you been shown?

RonL