# Answer on Algebra Question for Sam

Question #35172

-3a-4b+2c=28 a+3b-4c=-31 2a+3c=11

Expert's answer

Second equation:a+3b-4c=-31We multiply it by 2:

2a+6b-8c=-62

Find 2a from the third equation and substitute into the second equation:

11-3c+6b-8c=-62

6b-11c=-73

Find 6b from the previous equation:

6b=11c-73

We multiply it by 2:

12b-22c=-146

First equation:

-3a-4b+2c=28

Multiply it by -3:

9a+12b-6c=-84

substitute 6b from previous result into the above mentioned equation:

9a+22c-146-6c=-84

We get systems of 2 equations in 2 variables:

9a+16c=62

2a+3c=11;

Subtract from the first equation the second, multiplied by 4:

a+4c=18

2a+3c=11

Find a from the first equation:

a=18-4c

and replace it with the second equation:

2(18-4c)+3c=11

36-5c=11

5c=25

hence

c=5

a=18-4c=18-4*5=-2

Substitute solutions a,c found in the second equation of the system:

a+3b-4c=-31

-2+3b-20=-31

3b=22-31=-9

hence

b=-3

Finally answer is the following:

a=-2

b=-3

c=5

2a+6b-8c=-62

Find 2a from the third equation and substitute into the second equation:

11-3c+6b-8c=-62

6b-11c=-73

Find 6b from the previous equation:

6b=11c-73

We multiply it by 2:

12b-22c=-146

First equation:

-3a-4b+2c=28

Multiply it by -3:

9a+12b-6c=-84

substitute 6b from previous result into the above mentioned equation:

9a+22c-146-6c=-84

We get systems of 2 equations in 2 variables:

9a+16c=62

2a+3c=11;

Subtract from the first equation the second, multiplied by 4:

a+4c=18

2a+3c=11

Find a from the first equation:

a=18-4c

and replace it with the second equation:

2(18-4c)+3c=11

36-5c=11

5c=25

hence

c=5

a=18-4c=18-4*5=-2

Substitute solutions a,c found in the second equation of the system:

a+3b-4c=-31

-2+3b-20=-31

3b=22-31=-9

hence

b=-3

Finally answer is the following:

a=-2

b=-3

c=5

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