Question

What is the solution (a, c) to this system of linear equations?

2a – 3c = –6

a + 2c = 11

(negative StartFraction 76 over 7 EndFraction, StartFraction 17 over 7 EndFraction)
(3, –4)
(3, 4)
(StartFraction 87 over 7 EndFraction, negative StartFraction 5 over 7 EndFraction)

Answer 1

Answer:  The correct option is

(C) (3, 4).

Step-by-step explanation:  We are given to find the solution (a, c) to the following system of linear equations :

$2a-3c=-6~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\a+2c=11~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Multiplying equation (ii) by 2, we have

$2(a+2c)=2\times11\\\\\Rightarrow 2a+4c=22~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

Subtracting equation (i) from (iii), we get

$(2a+4c)-(2a-3c)=22-(-6)\\\\\Rightarrow 7c=28\\\\\Rightarrow c=\dfrac{28}{7}\\\\\Rightarrow c=4.$

From equation (i), we get

$2a-3\times4=-6\\\\\Rightarrow 2a=12-6\\\\\Rightarrow 2a=6\\\\\Rightarrow a=\dfrac{6}{2}\\\\\Rightarrow a=3.$

Thus, the required solution is (a, c) = (3, 4).

Option (C) is CORRECT.

Answer 2

The solution (a, c) to this system of linear equations is (3, 4)

System of equations

Given the system of equations expressed as:

2a – 3c = –6

a + 2c = 11

From equation 2; a = 11 - 2c

Substitute into equation 1 to have:

2(11-2c) - 3c = -6

22 - 4c - 3c = -6
22 - 7c = -6
-7c= -28

c = 4

Recall that a = 11 - 2c

a = 11 - 2(4)

a = 11 - 8

a = 3

Hence the solution (a, c) to this system of linear equations is (3, 4)

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