Question

12. Solve the equations x + y + z = 8, 3x + 5y - 7z=14, x
y + 2z = 6 by using Cramer's
method.​

Answer 1

z=11/7

y=20/7

x=47/7

hope it's help you.

Answer 2

Answer:

By using the Cramer's method, the solution of given three equations is $x=25,\;\; y=-15$ and $z=-2$.

Step-by-step explanation:

We have given the system of  three linear equations:

x+ y + z =8                                                        ................(1)

3x + 5y - 7z = 14                                                ................(2)

x + y + 2z = 6                                                     ................(3)

From the above three linear equations, we can find Δ:

$\triangle =\left[\begin{array}{ccc}1&1&1\\3&5&-7\\1&1&2\end{array}\right]$

$\triangle =1\left[\begin{array}{cc}5&-7\\1&2\end{array}\right]-1\left[\begin{array}{cc}3&-7\\1&2\end{array}\right]+1\left[\begin{array}{cc}3&5\\1&1\end{array}\right]$

$\triangle = 1 [(5)(2)-(1)(-7)] -1 [(3)(2)-(1)(-7)] +1 [(3)(1)-(5)(1)]$

Δ = 1 (10+7)-1 (6+7)+1 (3-5)

Δ =17- 13 -2

Δ =2

Since, Δ≠0 it means the system is consistent with unique solution and Cramer's method can be applied.

$\triangle_{x} =\left[\begin{array}{ccc}8&14&6\\3&5&-7\\1&1&2\end{array}\right]$

$\triangle _{x}=8(10+7)-1(28+42)+1(14-30)$

$\triangle_{x}=136-70-16$

$\triangle_{x}=50$

Now,

$\triangle_{y} =\left[\begin{array}{ccc}8&1&1\\14&5&-7\\6&1&2\end{array}\right]$

$\triangle_{y}=1(28+42)-8(6+7)+1(18-14)$

$\triangle_{y}= 70-104+4$

$\triangle_{y}=-30$

$\triangle_{z} =\left[\begin{array}{ccc}1&1&8\\3&5&14\\1&1&6\end{array}\right]$

$\triangle_{z}=1(30-14)-1(18-14)+8(3-5)$

$\triangle_{z}= 16-4-16$

$\triangle_{z}=-4$

Now, $\triangle_{x}=50,\;\; \triangle_{y}=-30\;\; and \;\;\triangle_{z}=-4$

$x=\frac{\triangle_{x}}{\triangle}=\frac{50}{2}=25$

$y=\frac{\triangle_{y}}{\triangle}=\frac{-30}{2}=-15$

$z=\frac{\triangle_{z}}{\triangle}=\frac{-4}{2}=-2$

Therefore, x=25, y=-15 and z=-2 is the solution of given equations.