In the following exercises, solve the system of equations by using substitution.

$\left\{\begin{array}{l}3x^2-y=0\\ y=2x-1\end{array}\right.$

We have given the following system of equations :-

$\left\{\begin{array}{l}3x^2-y=0\\ y=2x-1\end{array}\right.$

We have to solve this system of equations. That is we have to find pair for values of $xandy$that satisfies this system of equations.

We will substitute the value of $y$from second equation in first one. Than we have a quadratic equation in $x$. After that we will solve the equations, to get required values.

The given system of equations is :-

$\left\{\begin{array}{l}3x^2-y=0\\ y=2x-1\end{array}\right.$

Substitute value of $y=2x-1$in first equation $3x^2-y=0$, then we have :-

$$\begin{gathered} 3x^2-(2x-1)=0 \\ \Rightarrow 3x^2-2x+1=0 \end{gathered}$$

We have :-

$3x^2-2x+1=0$.

Here this does not factorized easily. So we check the discriminant.

The discriminant of a quadratic equation is $D=b^2-4ac$. Where $b$is coefficient of middle term and $aandc$are coefficients of first and last terms respectively.

Then the discriminant of above quadratic equation is :-

$$\begin{gathered} D={(-2)}^2-(4\times 3\times 1) \\ \Rightarrow D=4-12 \\ \Rightarrow D=-8 \end{gathered}$$

Here D is negative. That is there is no real solution of this quadratic equation.

So we can conclude that given system has no solution.