Step 1

Given: The system of equations is

2x+3y=4

4x-6y=8

Step 2

Solution:

Obtain the value of D as follows.

\(\displaystyle{D}{\left|\begin{array}{cc} {2}&{3}\\{4}&-{6}\end{array}\right|}\)

=2(-6)-(3)(4)

=-12-12

\(\displaystyle=-{24}\ne{0}\)

As D is not equal to zero, the unique solution exists.

Step 3

Obtain the value of \(\displaystyle{D}_{{1}}\) as follows.

\(\displaystyle{D}_{{1}}={\left|\begin{array}{cc} {4}&{3}\\{8}&-{6}\end{array}\right|}\)

=(4)(-6)-3(8)

=-24-24

=-48

Step 4

Obtain the value of \(\displaystyle{D}_{{2}}\) as follows.

\(\displaystyle{D}_{{2}}={\left|\begin{array}{cc} {2}&{4}\\{4}&{8}\end{array}\right|}\)

=2(8)-(4)(4)

=16-16

=0

Step 5

Thus, the value of x and y by Cramer’s rule becomes

\(\displaystyle{x}=\frac{{D}_{{1}}}{{D}}=\frac{{-{48}}}{{-{24}}}={2}\)

\(\displaystyle{y}=\frac{{D}_{{2}}}{{D}}=\frac{{0}}{{-{24}}}={0}\)

Step 6

Therefore, the solution of the system of equations is

x=2 and y =0

Given: The system of equations is

2x+3y=4

4x-6y=8

Step 2

Solution:

Obtain the value of D as follows.

\(\displaystyle{D}{\left|\begin{array}{cc} {2}&{3}\\{4}&-{6}\end{array}\right|}\)

=2(-6)-(3)(4)

=-12-12

\(\displaystyle=-{24}\ne{0}\)

As D is not equal to zero, the unique solution exists.

Step 3

Obtain the value of \(\displaystyle{D}_{{1}}\) as follows.

\(\displaystyle{D}_{{1}}={\left|\begin{array}{cc} {4}&{3}\\{8}&-{6}\end{array}\right|}\)

=(4)(-6)-3(8)

=-24-24

=-48

Step 4

Obtain the value of \(\displaystyle{D}_{{2}}\) as follows.

\(\displaystyle{D}_{{2}}={\left|\begin{array}{cc} {2}&{4}\\{4}&{8}\end{array}\right|}\)

=2(8)-(4)(4)

=16-16

=0

Step 5

Thus, the value of x and y by Cramer’s rule becomes

\(\displaystyle{x}=\frac{{D}_{{1}}}{{D}}=\frac{{-{48}}}{{-{24}}}={2}\)

\(\displaystyle{y}=\frac{{D}_{{2}}}{{D}}=\frac{{0}}{{-{24}}}={0}\)

Step 6

Therefore, the solution of the system of equations is

x=2 and y =0