Question

Solve each system of equations by graphing.

$\begin{array}{c}23.\frac{1}{2}x-y=0\\ \frac{1}{4}x+\frac{1}{2}y=-2\end{array}$

Short answer

The solution of system of equations is $\left(-4,-2\right)$.

Step-by-step solution

Step-1 – Apply the concept of slope-intercept form

Equation of line in slope intercept form is expressed below.

$y=mx+c$

Where m is the slope and c is the intercept of y-axis.

Step-2 –Write the equations in slope-intercept form

Consider the first equation $\frac{1}{2}x-y=0$.

Rewrite the equation in form of slope-intercept form.

Add the term y on both the sides.

$\begin{array}{c}\frac{1}{2}x-y+y=y\\ \frac{1}{2}x=y\\ y=\frac{1}{2}x+0\end{array}$

Now, the equation is in the form $y=mx+c$. Here slope m of the line is $\frac{1}{2}$ and intercept of y-axis c is 0.

Now, consider the second equation $\frac{1}{4}x+\frac{1}{2}y=-2$.

Rewrite the equation in form of slope-intercept form.

Subtract both sides by $\frac{1}{4}x$.

$\begin{array}{c}\frac{1}{4}x+\frac{1}{2}y-\frac{1}{4}x=-2-\frac{1}{4}x\\ \frac{1}{2}y=-2-\frac{1}{4}x\end{array}$

Next, multiply both sides by 2 and rearrange the terms.

$\begin{array}{c}y=-4-\frac{1}{2}x\\ y=-\frac{1}{2}x-4\end{array}$

Now, the equation is in the form $y=mx+c$. Here slope m of the line is $-\frac{1}{2}$ and intercept of y-axis c is $-4$.

Step-3 – Identify the point of intersection of the equations

Plot the equations on the same plane and the point where both the equations intersect is the solution of the system of the equations.

The red line denotes the equation $\frac{1}{2}x-y=0$and blue line denotes the equation $\frac{1}{4}x+\frac{1}{2}y=-2$.

Therefore, the point of intersection is $\left(-4,-2\right)$.

Step-4 – Verify that point satisfies system of equations

The point of intersection is solution of system of equations if the point satisfies both the equation.

Substitute the point $\left(-4,-2\right)$in the equation $\frac{1}{2}x-y=0$.

Substitute x as $-4$ and y as $-2$and check whether right hand side is equal to left hand side of the equation.

$\begin{array}{c}\frac{1}{2}\left(-4\right)-\left(-2\right)=0\\ -2+2=0\\ 0=0\end{array}$

Since, this is true so the point $\left(-4,-2\right)$satisfy the equation $\frac{1}{2}x-y=0$.

Substitute the point $\left(-4,-2\right)$in the equation $\frac{1}{4}x+\frac{1}{2}y=-2$.

Substitute x as $-4$ and y as $-2$and check whether right hand side is equal to left hand side of the equation.

$\begin{array}{c}\frac{1}{4}\left(-4\right)+\frac{1}{2}\left(-2\right)=-2\\ -1-1=-2\\ -2=-2\end{array}$

Since, this is true so the point $\left(-4,-2\right)$satisfy the equation .

Hence, the solution of the system of equations is .