In the following exercises, which ordered pair are solutions to the given equation :

$y=2x-5$

(a)$(0,-5)$(b)$(2,1)$(c)$\left(\frac{1}{2},-4\right)$

The given equation is $y=2x-5$and the three coordinates are $\left(0,-5\right),\left(2,1\right)$and $\left(\frac{1}{2},-4\right)$.

We need to find the solution of the given equation.

To check the first coordinate : $\left(0,-5\right)$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 $x=0$and $y=-5$.

From the equation : $$\begin{gathered} y=2x-5 \\ \Rightarrow 2x-y=5 \end{gathered}$$

The R.H.S is $5$, we need to solve the L.H.S :$$\begin{gathered} 2x-y=2\left(0\right)-(-5) \\ =5 \end{gathered}$$

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

So, it is the solution of the given equation.

To check the second coordinate : $\left(2,1\right)$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 $x=2$and $y=1$.

From the equation : $$\begin{gathered} y=2x-5 \\ \Rightarrow 2x-y=5 \end{gathered}$$

The R.H.S is $5$, we need to solve the L.H.S :

$$\begin{gathered} 2x-y=2\left(2\right)-1 \\ =4-1 \\ =3 \end{gathered}$$

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

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

To check the third coordinate : $\left(\frac{1}{2},-4\right)$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 $x=\frac{1}{2}$and $y=-4$.

From the equation : $$\begin{gathered} y=2x-5 \\ \Rightarrow 2x-y=5 \end{gathered}$$

The R.H.S is $5$, we need to solve the L.H.S :

$$\begin{gathered} 2x-y=2\left(\frac{1}{2}\right)-(-4) \\ =1+4 \\ =5 \end{gathered}$$

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

So, it is the solution of the given equation.