Question

Solve the system of equations 3x+4y+5z=a; 4x+5y+6z=b; 5x+6y+7z=c; possess a solution if a+c=2b. Solve when (a, b,



c.=(1, 2,3)

Answer 1

Show that the equations 3x + 4y + 5z = a, 4x + 5y + 6z = b, 5x + 6y + 7z = c do not have a solution unless a + c = 2b.

Answer 2

The solution of the given system of equations is $\left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}0\\4\\-3\end{array}\right]$

Step-by-step explanation:

Given: The system of equations

$3x+4y+5z=a\\4x+5y+6z=b\\5x+6y+7z=c$ and (a + c = 2b)

To Find: Solution of the given system of equations for (a,b,c) = (1,2,3)

Solution:

• Finding the solution of the given system of equations

For the given equation, we can write them in the form of a matrix such that AX = B;

$\left[\begin{array}{ccc}3&4&5\\4&5&6\\5&6&7\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}a\\b\\c\end{array}\right]$

Since we are given that the solution of possible if a+c=2b, and for the given values (a,b,c) = (1,2,3), the following condition is satisfied. Therefore, solution of the system of equations at (a,b,c) = (1,2,3) can be found as;

$\left[\begin{array}{ccc}3&4&5\\4&5&6\\5&6&7\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\2\\3\end{array}\right]$

using row operation $R_1 \rightarrow R_2 - R_1$

$\Rightarrow \left[\begin{array}{ccc}1&1&1\\4&5&6\\5&6&7\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\2\\3\end{array}\right]$

using row operation $R_2 \rightarrow R_2 - 4R_1$

$\Rightarrow \left[\begin{array}{ccc}1&1&1\\0&1&2\\5&6&7\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\-2\\3\end{array}\right]$

using row operation $R_3 \rightarrow R_3 - 5R_1$

$\Rightarrow \left[\begin{array}{ccc}1&1&1\\0&1&2\\0&1&2\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\-2\\-2\end{array}\right]$

using row operation $R_3 \rightarrow R_3 - R_3/R_2$

$\Rightarrow \left[\begin{array}{ccc}1&1&1\\0&1&2\\0&0&1\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\-2\\-3\end{array}\right]$

The above matrix can be rewritten into a system of equations and we can find values of x, y, and z as follows,

$z = -3$

$y + 2z = -2 \Rightarrow y = 4$

$x+y+z=1 \Rightarrow x = 0$

Hence, the solution of the system of the given equations is $\left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}0\\4\\-3\end{array}\right]$