Verified Answer :

calculista

Answer:

option C

\((3,4)\)

Step-by-step explanation:

we have

\(2a-3c=-6\) -----> equation A

\(a+2c=11\) -----> equation B

Multiply the equation B by \(-2\)

\(-2(a+2c)=-2*11\)

\(-2a-4c=-22\) ------> equation C

Adds equation A and equation C

\(2a-3c=-6\\-2a-4c=-22\\----------\\-3c-4c=-6-22\\-7c=-28\\c=4\)

Find the value of a

substitute the value of c in equation A

\(2a-3(4)=-6\)

\(2a=6\)

\(a=3\)

The solution is the point \((3,4)\)

Answer:

option c is correct.

3, 4

Step-by-step explanation:

Given the system of equations:

\(2a-3c = -6\)          .....[1]

\(a+2c = 11\)             .....[2]

Multiply equation [2] by 2 both sides we have;

\(2a + 4c= 22\)            ......[3]

Subtract equation [2] from [3] we have;

\(2a+4c - 2a+3c = 22 - (-6)\)

⇒\(4c+3c = 28\)

Combine like terms;

\(7c = 28\)

Divide both sides by 7 we have;

c = 4

Substitute this in [1] we have;

2a- 3(4) = -6

2a -12 = -6

Add 12 to both sides we have;

2a = 6

Divide both sides by 3 we have;

a = 3

Therefore, the solution to this system of linear equations is, (3, 4)

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