Question

Solve each system of equations by graphing.

49.

$\begin{array}{l}\mathbf{x}+\mathbf{y}=\mathbf{7}\\ \frac{1}{2}\mathbf{x}-\mathbf{y}=-\mathbf{1}\end{array}$

Short answer

The solution of system of equations is $\mathbf{x}\mathbf{=}\mathbf{4}$ and $\mathbf{y}\mathbf{=}\mathbf{3}$.

Step-by-step solution

­- Description of step.

Number the given system of equations.

$\begin{array}{l}x+y=7\left(1\right)\\ \frac{1}{2}x-y=-1\left(2\right)\end{array}$

­- Description of step.

The first equation is: $x+y=7$.

When $x=0$,

$\begin{array}{l}0+y=7\\ y=7\end{array}$

Therefore, when $x=0$, $y=7$.

When $x=-1$,

$\begin{array}{l}-1+y=7\\ y=7+1\\ y=8\end{array}$

Therefore, when $x=-1$, $y=8$

Therefore, the equation $x+y=7$ is the equation of the line passing through the points $\left(0,7\right)$ and $\left(-1,8\right)$.

­- Graph the equation x+y=7.

The graph of the equation $x+y=7$ is:

­- Description of step.

The second equation is: $\frac{1}{2}x-y=-1$.

When $x=0$,

$\begin{array}{l}\frac{1}{2}\left(0\right)-y=-1\\ 0-y=-1\\ -y=-1\\ y=1\end{array}$

Therefore, when $x=0$, $y=1$.

When $x=-2$,

$\begin{array}{l}\frac{1}{2}\left(-2\right)-y=-1\\ -1-y=-1\\ -y=-1+1\\ -y=0\\ y=0\end{array}$

Therefore, when $x=-2$, $y=0$

Therefore, the equation $\frac{1}{2}x-y=-1$ is the equation of the line passing through the points $\left(0,1\right)$ and $\left(-2,0\right)$.

­- Graph the equation 12x-y=-1.

The graph of the equation $\frac{1}{2}x-y=-1$ is:

­- Graph the equations x+y=7 and 12x-y=-1.

The graph of the equations $x+y=7$ and $\frac{1}{2}x-y=-1$ is:

­- Description of step.

From the graph of the equations $x+y=7$ and $\frac{1}{2}x-y=-1$, it can be noticed that the lines are intersecting at a point $\left(4,3\right)$.

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

Therefore, the solution of the given system of equations is $\left(4,3\right)$.

That implies, the solution of the given system of equation is $x=4$ and $y=3$.