Question

in exercises 1 through 10, find all solutions of the linear systems using elimination.Then check your solutions.

3.$$\begin{gathered} \left|2x+4y=3 \\ 3x+6y=2\right| \end{gathered}$$

Short answer

The given system of equations has no solution.

Step-by-step solution

 Step 1:Transforming the system

To get the solution, we will transform the value of x,y

$$\begin{gathered} \left|\mathrm{2}\mathrm{x}\mathrm{+}\mathrm{4}\mathrm{y}\mathrm{=}\mathrm{3} \\ \mathrm{3}\mathrm{x}\mathrm{+}\mathrm{6}\mathrm{y}\mathrm{=}\mathrm{2}\right| \end{gathered}$$Into the form$$\begin{gathered} \left|x=.... \\ y=.... \\ \right| \end{gathered}$$

Eliminating the variables

In the given system of equations, we can eliminate the variables by adding or subtracting the equations.

In this system, we can eliminate the variable x from equation 1. First of all multiply the first equation by 3 and multiply the second equation by 2.

$$\begin{gathered} \left|6x+12y=9 \\ 6x+12y=4\right| \end{gathered}$$

Now subtract the second equation from the first equation.

$$\begin{gathered} \left|6x-6x+12y-12y=9-4 \\ 6x+12y=4\right|=\left|0=5 \\ 6x+12y=4\right| \end{gathered}$$

Now we can for the any value of x,y the 0 will never be equal to 5.

Hence, the given system of equation has no solution.