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Chapter 5: Q73. (page 228)

Solve the following system of equations?

a+b+c=62a鈭�b+3c=16a+3b鈭�2c=鈭�6

Short Answer

Expert verified

The solution for the system of equations isa=2,b=0,c=4 .

Step by step solution

01

Step 1. Write down the given information.

The given system of equations are,

a+b+c=6....(1)2a鈭�b+3c=16....(2)a+3b鈭�2c=鈭�6....(3)

02

Step 2. Calculation.

Multiply (1) by 2 and subtract from (2) gives,

(2a鈭�b+3c)鈭�(2a+2b+2c)=16鈭�12(2a鈭�b+3c)鈭�(2a+2b+2c)=4鈭�3b+c=4....(4)

Subtract (3) from (1) gives,

(a+b+c)鈭�(a+3b鈭�2c)=6+6鈭�2b+3c=12鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌�....(5)

Multiply (4) by 3 and subtract from (5) gives,

(鈭�2b+3c)鈭�(鈭�9b+3c)=12鈭�127b=0b=0

Plugging b=0 in (4) gives c=4.

Again plugging b=0 and c=4in (1),

a+b+c=6....(From(1))a+(0)+(4)=6....(Pluggingb=0,c=4)a=2

03

Step 3. Conclusion.

The solution for the system of equations is a=2,b=0,c=4.

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