To calculate: The solution for the system of provided equations: 5a−2b+3c=10 −3a+b−2c=−7 a+4b−4c=−3

(A) ${\displaystyle 5a-2b+3c=10}$
(B) ${\displaystyle -3a+b-2c=-7}$
(C) ${\displaystyle a+4b-4c}$ ${\displaystyle -3}$
Consider (A) or (B):
(A) ${\displaystyle 5a-2b+3c=10}$
(B) ${\displaystyle -3a+b-2c=-7}$
Multiply (B) with 2:
(A) ${\displaystyle 5a-2b+3c=10}$
2 ${\displaystyle \cdot }$ (B) - 6a + 2b - 4c = -14
Now, add (A) and (B):
(D) ${\displaystyle -a-c=-4}$
Consider (A) and (C):
(A) ${\displaystyle 5a-2b+3c=10}$
(C) ${\displaystyle a+4b-4c=-3}$
Multiply (A) with 2:
2 ${\displaystyle \cdot }$ (A) 10a - 4b + 6c = 20
(C) ${\displaystyle a+4b-4c=-3}$
Add both the above equastion:
(E) ${\displaystyle 11a+2c=17}$
Now consider (D) and (E):
(D) ${\displaystyle -a-c=-4}$
(E) ${\displaystyle 11a+2c=17}$
Multipy (D) with 2:
2 ${\displaystyle \cdot }$ (D) - 2a - 2c = -8
(E) ${\displaystyle 11a+2c=17}$
Now, add both the equastions and solve for a:
${\displaystyle 9a=9}$
(dividing both sides by 9)
${\displaystyle a=1}$
Substitute 1 for a in (D):
${\displaystyle -\left(1\right)-c=-4}$
Adding 1 to both sides and then dividing by -1,
${\displaystyle -c=-3}$
${\displaystyle c=3}$
Substitude 3 for c and 1 for a in (A) and solve for b:
${\displaystyle 5\left(1\right)-2b+3\left(3\right)=10}$
${\displaystyle -2b+14=10}$
Substracting 14 from both sides and then dividing by -2,
${\displaystyle -2b=-4}$
${\displaystyle b=2}$
${\displaystyle -2b=-4}$
${\displaystyle b=2}$
So, the values obtained are a=1, b=2 and c=3. Substitute thease values
in each of the provided equations to verify:
First Equation: ${\displaystyle 5\left(1\right)-2\left(2\right)+3\left(3\right)}$ $$ 10
${\displaystyle 5-4+9}$ $$ 10
10 $$ 10
The result is true.
Second Equation:
-3 (1) + (2) - 2(3) $$ -7
-3 + 2 -6 $$ -7
-7 $$ -7
The result is true.
Third Equation:
(1) + 4 (2) - 4(3) $$ -3
1 + 8 - 12 $$ -3
-3 $$ -3
The result is true.
5a - 2b + 3c = 10
Therefore, the solution of the system of equations ${\displaystyle -3a+b-2c=-7}$ is
${\displaystyle a+4b-4c=-3}$