Chapter 3: Q27. (page 148)

Solve each system of equations.
$\begin{matrix} 2 a + b - c = 5 \\ a - b + 3 c = 9 \\ 3 a - 6 c = 6 \end{matrix}$

Short Answer

Expert verified

The solution is $a = 4, b = - 2, c = 1$ .

Step by step solution

Step-1 – Use elimination to make a system of two equations in two variables.

Given equations are-:

$\begin{matrix} 2 a + b - c = 5....... (1) \\ a - b + 3 c = 9....... (2) \\ 3 a - 6 c = 6....... (3) \end{matrix}$

Adding first and second equation, we get

$\begin{matrix} 2 a + a - c + 3 c = 5 + 9 \\ 3 a + 2 c = 14...... (4) \end{matrix}$

Step-2 – Solve the system of two equations.

$\begin{matrix} 3 a - 6 c = 6....... (3) \\ 3 a + 2 c = 14........ (4) \end{matrix}$

Subtracting fourth equation from third equation, we get

$\begin{matrix} - 6 c - 2 c = 6 - 14 \\ - 8 c = - 8 \\ c = 1 \end{matrix}$

Putting the value of $c$ in equation - $(3)$ , we get

$\begin{matrix} 3 a - 6 c = 6 \\ 3 a - 6 (1) = 6 \\ 3 a = 6 + 6 \\ 3 a = 12 \\ a = 4 \end{matrix}$

Step-3 – Substituting the value according to the coordinates in first equation.

$\begin{matrix} 2 a + b - c = 5 \\ 2 (4) + b - 1 = 5 \\ 8 + b - 1 = 5 \\ b = 5 - 7 \\ b = - 2 \end{matrix}$

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Solve each system of equations.
$\begin{matrix} 2 a + b - c = 5 \\ a - b + 3 c = 9 \\ 3 a - 6 c = 6 \end{matrix}$

Short Answer

Step by step solution

Step-1 – Use elimination to make a system of two equations in two variables.

Step-2 – Solve the system of two equations.

Step-3 – Substituting the value according to the coordinates in first equation.

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Solve each system of equations.2a+b-c=5a-b+3c=93a-6c=6

Short Answer

Step by step solution

Step-1 – Use elimination to make a system of two equations in two variables.

Step-2 – Solve the system of two equations.

Step-3 – Substituting the value according to the coordinates in first equation.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

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Study anywhere. Anytime. Across all devices.

Solve each system of equations.
$\begin{matrix} 2 a + b - c = 5 \\ a - b + 3 c = 9 \\ 3 a - 6 c = 6 \end{matrix}$