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

$y=\frac{1}{3}x+1$

(a)$\left(-3,0\right)$(b)$\left(9,4\right)$(c)$\left(-6,-1\right)$

The given equation is $y=\frac{1}{3}x+1$and the three coordinates are $\left(-3,0\right),\left(9,4\right)$and $\left(-6,-1\right)$.

We need to solve the equation.

To check the first coordinate : $\left(-3,0\right)$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 $x=-3$and $y=0$.

From the equation : $$\begin{gathered} y=\frac{1}{3}x+1 \\ \Rightarrow \frac{1}{3}x-y=-1 \end{gathered}$$

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

$$\begin{gathered} \frac{1}{3}x-y=\frac{1}{3}(-3)-0 \\ =-1 \end{gathered}$$

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

So, it is the solution of the given equation .

To check the second coordinate : $\left(9,4\right)$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 $x=9$and $y=4$.

From the equation : $$\begin{gathered} y=\frac{1}{3}x+1 \\ \Rightarrow \frac{1}{3}x-y=-1 \end{gathered}$$

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

$$\begin{gathered} \frac{1}{3}x-y=\frac{1}{3}\left(9\right)-4 \\ =3-4 \\ =-1 \end{gathered}$$

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

So, it is the solution of the given equation.

To check the third coordinate : $\left(-6,-1\right)$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 $x=-6$and $y=-1$.

From the equation : $$\begin{gathered} y=\frac{1}{3}x+1 \\ \Rightarrow \frac{1}{3}x-y=-1 \end{gathered}$$

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

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

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

So, it is the solution of the given equation.