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4 votes
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

User Yardena
by
8.0k points

2 Answers

7 votes

Answer:

option c is correct.

3, 4

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)

User Lonami
by
7.8k points
3 votes

Answer:

option C


(3,4)

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)

User Pnavk
by
7.3k points

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