5A-17B+10C=0

11A-5B-C=280

2A+20B-27C=40

Solve for A,B,C

asked by guest
on Sep 19, 2024 at 9:11 pm



You asked:

Given \( 5 A - 17 B + 10 C = 0 \), \( 11 A - 5 B - C = 280 \), \( 2 A + 20 B - 27 C = 40 \), evaluate \( A \), \( B \), \( C \).

MathBot Answer:

The system of linear equations has exactly one solution.

The substitutions are \( A = 36 \), \( B = 20 \), \( C = 16 \).

\[A = 36\]

\[B = 20\]

\[C = 16\]

Solve \(5 A - 17 B + 10 C = 0\) for \(A\). \[A = \frac{17 B}{5} - 2 C\]

Substitute \(\frac{17 B}{5} - 2 C\) for \(A\) in each of the remaining equations and simplify. $$\begin{aligned}11 A - 5 B - C &= 280 \\ 11 \left(\frac{17 B}{5} - 2 C\right) - 5 B - C &= 280 \\ 162 B - 115 C &= 1400 \end{aligned}$$$$\begin{aligned}2 A + 20 B - 27 C &= 40 \\ 2 \left(\frac{17 B}{5} - 2 C\right) + 20 B - 27 C &= 40 \\ 134 B - 155 C &= 200 \end{aligned}$$

Solve \(162 B - 115 C = 1400\) for \(B\). \[B = \frac{115 C}{162} + \frac{700}{81}\]

Substitute \(\frac{115 C}{162} + \frac{700}{81}\) for \(B\) in \(134 B - 155 C = 200\) and simplify. $$\begin{aligned}134 B - 155 C &= 200 \\ 134 \left(\frac{115 C}{162} + \frac{700}{81}\right) - 155 C &= 200 \\ C &= 16 \end{aligned}$$

Use substitution of the numerical value of \(C\) to get the values of \(A\) and \( B\). $$\begin{aligned}B &= \frac{115 C}{162} + \frac{700}{81} \\ B &= \frac{115 \cdot 16}{162} + \frac{700}{81} \\ B &= 20\end{aligned}$$$$\begin{aligned}A &= \frac{17 B}{5} - 2 C \\ A &= \frac{17 \cdot 20}{5} - 2 \cdot 16 \\ A &= 36\end{aligned}$$

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