Question

Solving the following system of equations for $a,b,c$

$3a-2b+c=6$
$a+b-c=1$
$a-b+2c=12$

• A. $a=3,b=-3,c=-8$
• B. $a=3,b=5,c=7$
• C. $a=3,b=5,c=-7$
• D. $a=7,b=3,c=-5$

Answer 1

The correct option is B $a=3,b=5,c=7$

To solve the equation we want to try to cancel out some of the letters to reduce the equations. We can do this by adding different equations to each other. If we add:

$3a-2b+c=6$
$+a+b-c=1$
We get a new equation,
$4a-b=7$
We can also add
$a+b-c=1$
(+a-b+2c = 12\)
to get $2a+c=13$
So we can rearrange the new equations to get
$b=4a-7;c=13-2a$
and sunstitute these into the original second line equation:
$a+(4a+7)-(13-2a)=1\to 7a-20=1$
$7a=21\to a=3$
Then we can solve $b$ and $c$:

$b=4.3-7=5$
$c=13-2.3=7$