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arias333

arias333

17.09.2021 •

Mathematics

0=A+B
1=A-3B+C
-1=-2A-2C

Find the values of A, B and C

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Ответ:

djfluffyman999

17.09.2021

Solving Systems of 3 Equations

$A = \frac16\\$ , $B = -\frac16\\$ and $C = \frac13\\$

Step-by-step explanation:

Given:

$&0 = A +B \\ &1 = A -3B +C \\ &-1 = 2A -2C$

Rewriting 0=A+B :

$0 = A +B \\ -A = B$

Rewriting 1=A-3B+C :

$1 = A -3B +C \\ 1 = A -3(-A) +C \\ 1 = A +3A +C \\ 1 = 4A +C \\ 1 -C = 4A \\ \frac{1 -C}{4} = A$

Rewriting -1=-2A-2C :

$-1 = -2A -2C \\ 2C -1 = -2A \\ \frac{2C -1}{-2} = A \\ -\frac{2C -1}{2} = A$

Since $A = -\frac{2C -1}{2}\\$ and $A = \frac{1 -C}{4}\\$ also, therefore $\frac{1 -C}{4} = -\frac{2C -1}{2}\\$ . We can now solve for from the resulting equation.

Solving for :

$\frac{1 -C}{4} = -\frac{2C -1}{2} \\ 1 -C = 2 \cdot -(2C -1) \\ 1 -C = -2(2C -1) \\ 1 -C = -4C +2 \\ -C = -4C +1 \\ 3C = 1 \\ C = \frac{1}{3}$

Solving for :

$\frac{1 -C}{4} = A \\ \frac{1 -\frac{1}{3}}{4} = A \\ \frac{\frac{3}{3} -\frac{1}{3}}{4} = A \\ \frac{\frac{2}{3}}{4} = A \\ \frac{2}{3} \cdot \frac{1}{4} = A \\ \frac{2}{12} = A \\ \frac16 = A$

Solving for :

$-A = B \\ -\frac16 = B$

0,0(0 оценок)

Ответ:

desi102

08.11.2021

Your answer should be C which is the third option

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