91Ó°ÊÓ

We are given an equation : $y=4x+3$and three coordinates : $\left(4,3\right),\left(-1,-1\right)$and $\left(\frac{1}{2},5\right)$.

We need to find the solution of the given equation.

To check the first coordinate : $(4,3)$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 $y=3$.

From the equation :

$$\begin{gathered} y=4x+3 \\ ⇒y-4x=3 \end{gathered}$$

The R.H.S is $3$, we need to solve L.H.S :

$$\begin{gathered} y-4x=3-4\left(4\right) \\ =3-16 \\ =-13 \end{gathered}$$

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

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

To check the second coordinate : $(-1,-1)$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 $x=-1$and $y=-1$.

From the equation : $$\begin{gathered} y=4x+3 \\ ⇒y-4x=3 \end{gathered}$$

The R.H.S is $3$, we need to solve L.H.S :

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

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

So, it is the solution of the given equation.

To check the third coordinate : $\left(\frac{1}{2},5\right)$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 $x=\frac{1}{2}$and $y=5$.

From the equation : $$\begin{gathered} y=4x+3 \\ ⇒y-4x=3 \end{gathered}$$

The R.H.S is $3$, we need to solve the L.H.S :

$$\begin{gathered} y-4x=5-4\left(\frac{1}{2}\right) \\ =5-2 \\ =3 \end{gathered}$$

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

So, it is a solution of the given equation.