In the following exercises, which ordered pairs are solutions to the given equations :

$y=\frac{1}{2}x-1$

(a) $\left(2,0\right)$(b)$\left(-6,-4\right)$(c)$\left(-4,-1\right)$

The given equation is $y=\frac{1}{2}x-1$and the three coordinates are $\left(2,0\right),\left(-6,-4\right)$$\left(-4,-1\right)$.

We need to solve the equation.

To check the first coordinate : $\left(2,0\right)$we need to put the values of $x$and $y$in the equation and see if L.H.S=R.H.S .

Here, we know that $x=2$and $y=0$.

From the equation : $$\begin{gathered} y=\frac{1}{2}x-1 \\ \Rightarrow \frac{1}{2}x-y=1 \end{gathered}$$

The R.H.S is $1$, we need to solve the L.H.S :

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

Therefore, L.H.S=R.H.S .

So, it is the solution of the given equation.

To check the second coordinate : $\left(-6,-4\right)$we need to put the values of $x$and $y$in the equation and see if L.H.S=R.H.S .

Here, we know that $x=-6$and $y=-4$.

From the equation : $$\begin{gathered} y=\frac{1}{2}x-1 \\ \Rightarrow \frac{1}{2}x-y=1 \end{gathered}$$

The R.H.S is $1$, we need to solve the L.H.S :

$$\begin{gathered} \frac{1}{2}x-y=\frac{1}{2}(-6)-(-4) \\ =-3+4 \\ =1 \end{gathered}$$

Therefore, L.H.S=R.H.S .

So, it is the solution of the given equation.

To check the third coordinate : $\left(-4,-1\right)$we need to put the values of $x$and $y$in the equation and see if L.H.S=R.H.S.

Here, we know that $x=-4$and $y=-1$.

From the equation : $$\begin{gathered} y=\frac{1}{2}x-1 \\ \Rightarrow \frac{1}{2}x-y=1 \end{gathered}$$

The R.H.S is $1$, we need to solve the L.H.S :

$$\begin{gathered} \frac{1}{2}x-y=\frac{1}{2}(-4)-(-1) \\ =-2+1 \\ =-1 \end{gathered}$$

Therefore, L.H.S is not equal to R.H.S .

So, it is not a solution of the given equation.