kolonelyf4

2021-11-17

To calculate: The solution for the system of provided equations:

$5a-2b+3c=10$

$-3a+b-2c=-7$

$a+4b-4c=-3$

Melinda Olson

Beginner2021-11-18Added 20 answers

(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

Now, add (A) and (B):

(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)$a+4b-4c=-3$

Add both the above equastion:

(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

$a+4b-4c=-3$

(B)

(C)

Consider (A) or (B):

(A)

(B)

Multiply (B) with 2:

(A)

2

Now, add (A) and (B):

(D)

Consider (A) and (C):

(A)

(C)

Multiply (A) with 2:

2

(C)

Add both the above equastion:

(E)

Now consider (D) and (E):

(D)

(E)

Multipy (D) with 2:

2

(E)

Now, add both the equastions and solve for a:

(dividing both sides by 9)

Substitute 1 for a in (D):

Adding 1 to both sides and then dividing by -1,

Substitude 3 for c and 1 for a in (A) and solve for b:

Substracting 14 from both sides and then dividing by -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:

10

The result is true.

Second Equation:

-3 (1) + (2) - 2(3)

-3 + 2 -6

-7

The result is true.

Third Equation:

(1) + 4 (2) - 4(3)

1 + 8 - 12

-3

The result is true.

5a - 2b + 3c = 10

Therefore, the solution of the system of equations