 # To calculate: The solution for the system of provided equations: kolonelyf4 2021-11-17 Answered
To calculate: The solution for the system of provided equations:
$5a-2b+3c=10$
$-3a+b-2c=-7$
$a+4b-4c=-3$
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(A) $5a-2b+3c=10$
(B) $-3a+b-2c=-7$
(C) $a+4b-4c$ $-3$
Consider (A) or (B):
(A) $5a-2b+3c=10$
(B) $-3a+b-2c=-7$
Multiply (B) with 2:
(A) $5a-2b+3c=10$
2 $\cdot$ (B) - 6a + 2b - 4c = -14
(D) $-a-c=-4$
Consider (A) and (C):
(A) $5a-2b+3c=10$
(C) $a+4b-4c=-3$
Multiply (A) with 2:
2 $\cdot$ (A) 10a - 4b + 6c = 20
(C) $a+4b-4c=-3$
(E) $11a+2c=17$
Now consider (D) and (E):
(D) $-a-c=-4$
(E) $11a+2c=17$
Multipy (D) with 2:
2 $\cdot$ (D) - 2a - 2c = -8
(E) $11a+2c=17$
Now, add both the equastions and solve for a:
$9a=9$
(dividing both sides by 9)
$a=1$
Substitute 1 for a in (D):
$-\left(1\right)-c=-4$
Adding 1 to both sides and then dividing by -1,
$-c=-3$
$c=3$
Substitude 3 for c and 1 for a in (A) and solve for b:
$5\left(1\right)-2b+3\left(3\right)=10$
$-2b+14=10$
Substracting 14 from both sides and then dividing by -2,
$-2b=-4$
$b=2$
$-2b=-4$
$b=2$
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: $5\left(1\right)-2\left(2\right)+3\left(3\right)$  10
$5-4+9$  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 $-3a+b-2c=-7$ is
$a+4b-4c=-3$