Q50. - Algebra 2

Question

Solve each system of equations by graphing.

50.

$\begin{array}{l}\mathbf{x}-\mathbf{2}\mathbf{y}=\mathbf{10}\\ \mathbf{2}\mathbf{x}-\mathbf{4}\mathbf{y}=\mathbf{12}\end{array}$

Step-by-Step Solution

Answer

There is no solution of the given system of equations.

Step 1 ­- Description of step.

Number the given system of equations.

$\begin{array}{c}x-2y=10\left(1\right)\\ 2x-4y=12\left(2\right)\end{array}$

Step 2 ­- Description of step.

The first equation is: $x-2y=10$.

When $x=0$,

$\begin{array}{c}0-2y=10\\ -2y=10\\ y=-5\end{array}$

Therefore, when $x=0$, $y=-5$.

When $x=2$,

$\begin{array}{c}2-2y=10\\ -2y=8\\ y=-4\end{array}$

Therefore, when $x=2$, $y=-4$

Therefore, the equation $x-2y=10$ is the equation of the line passing through the points $\left(0,-5\right)$ and $\left(2,-4\right)$.

Step 3 ­- Graph the equation x - 2 y = 10 .

The graph of the equation $x-2y=10$ is:

Step 4 ­- Description of step.

The second equation is: $2x-4y=12$.

When $x=0$,

$\begin{array}{c}2x-4y=12\\ 2\left(0\right)-4y=12\\ 0-4y=12\\ -4y=12\\ y=-3\end{array}$

Therefore, when $x=0$, $y=-3$.

When $x=2$,

$\begin{array}{c}2\left(2\right)-4y=12\\ 4-4y=12\\ -4y=8\\ y=-2\end{array}$

Therefore, when $x=2$, $y=-2$

Therefore, the equation $2x-4y=12$ is the equation of the line passing through the points $\left(0,-3\right)$ and $\left(2,-2\right)$.

Step 5 ­- Graph the equation 2 x - 4 y = 12 .

The graph of the equation $2x-4y=12$ is:

Step 6 ­- Graph the equations x - 2 y = 10 and 2 x - 4 y = 12 .

The graph of the equations $x-2y=10$ and $2x-4y=12$ is:

Step 7 ­- Description of step.

From the graph of the equations $x-2y=10$ and $2x-4y=12$, it can be noticed that the lines are parallel to  each other and therefore there is no point of intersection of these lines.

The solution of the system of linear equations in two variables is given by the intersection point of the lines.

Therefore, there is no solution of the given system of equations.