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What is the solution to this system of linear equations?

2a – 3c = –6

a + 2c = 11

(a) -76/7, 17/7
(b) 3,-4
(c) 3/4
(d) 87/7, -5/7

Respuesta :

calculista

Answer:

option C

[tex](3,4)[/tex]

Step-by-step explanation:

we have

[tex]2a-3c=-6[/tex] -----> equation A

[tex]a+2c=11[/tex] -----> equation B

Multiply the equation B by [tex]-2[/tex]

[tex]-2(a+2c)=-2*11[/tex]

[tex]-2a-4c=-22[/tex] ------> equation C

Adds equation A and equation C

[tex]2a-3c=-6\\-2a-4c=-22\\----------\\-3c-4c=-6-22\\-7c=-28\\c=4[/tex]

Find the value of a

substitute the value of c in equation A

[tex]2a-3(4)=-6[/tex]

[tex]2a=6[/tex]

[tex]a=3[/tex]

The solution is the point [tex](3,4)[/tex]

Answer:

option c is correct.

3, 4

Step-by-step explanation:

Given the system of equations:

[tex]2a-3c = -6[/tex]          .....[1]

[tex]a+2c = 11[/tex]             .....[2]

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

[tex]2a + 4c= 22[/tex]            ......[3]

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

[tex]2a+4c - 2a+3c = 22 - (-6)[/tex]

⇒[tex]4c+3c = 28[/tex]

Combine like terms;

[tex]7c = 28[/tex]

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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