What is a+b+c=0,a-3b+c=1,-2a-1c=-1 ?

The solution to a+b+c=0,a-3b+c=1,-2a-1c=-1 is a= 3/4 ,b=-1/4 ,c=-1/2

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a+b+c=0,a-3b+c=1,-2a-1c=-1

a + b + c = 0, a - 3 b + c = 1, - 2 a - 1 c = - 1

a = \frac{3}{4}, b = - \frac{1}{4}, c = - \frac{1}{2}

Solution steps

a + b + c = 0 a - 3 b + c = 1 - 2 a - 1 \cdot c = - 1

Isolate

a

for

- 2 a - 1 c = - 1 : a = - \frac{- 1 + c}{2}

Substitute

a = - \frac{- 1 + c}{2}

[- \frac{- 1 + c}{2} + b + c = 0 - \frac{- 1 + c}{2} - 3 b + c = 1]

Simplify

[b + \frac{1 + c}{2} = 0 - 3 b + \frac{1 + c}{2} = 1]

Isolate

b

for

b + \frac{1 + c}{2} = 0 : b = - \frac{c + 1}{2}

Substitute

b = - \frac{c + 1}{2}

[- 3 (- \frac{c + 1}{2}) + \frac{1 + c}{2} = 1]

Simplify

[2 (c + 1) = 1]

Isolate

c

for

2 (c + 1) = 1 : c = - \frac{1}{2}

For

b = - \frac{c + 1}{2}

Substitute

c = - \frac{1}{2}

b = - \frac{- \frac{1}{2} + 1}{2}

- \frac{- \frac{1}{2} + 1}{2} = - \frac{1}{4}

b = - \frac{1}{4}

For

a = - \frac{- 1 + c}{2}

Substitute

b = - \frac{1}{4}, c = - \frac{1}{2}

a = - \frac{- 1 - \frac{1}{2}}{2}

- \frac{- 1 - \frac{1}{2}}{2} = \frac{3}{4}

a = \frac{3}{4}

The solutions to the system of equations are:

a = \frac{3}{4}, b = - \frac{1}{4}, c = - \frac{1}{2}

2 x - 5 y = 14, 3 x + y = 4

x = 6, y = 3

5 x + 3 y = 10, 2 x + \frac{6}{5} y = 4

71 x + 37 y = 25, 337 x + 71 y = 287

a = 5, b = 4

What is a+b+c=0,a-3b+c=1,-2a-1c=-1 ?
The solution to a+b+c=0,a-3b+c=1,-2a-1c=-1 is a= 3/4 ,b=-1/4 ,c=-1/2