In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

$$\begin{gathered} ⓐ\left\{\begin{array}{l}8x-15y=-32\\ 6x+3y=-5\end{array}\right. \\ ⓑ\left\{\begin{array}{l}x=4y-3\\ 4x-2y=-6\end{array}\right. \end{gathered}$$

The given equations are as:

$$\begin{gathered} \left\{\begin{array}{l}8x-15y=-32\\ 6x+3y=-5\end{array}\right. \\ \left\{\begin{array}{l}x=4y-3\\ 4x-2y=-6\end{array}\right. \end{gathered}$$

In a) part equations are

$\left\{\begin{array}{l}8x-15y=-32\\ 6x+3y=-5\end{array}\right.$

The above 2 equations are in the standard form and if we will multiply the 2nd equation by 5 and then just add the two linear equations, it will be very easy to solve.

Therefore Elimination will be best to solve the part a.

We are given system of linear equation,

$\left\{\begin{array}{l}x=4y-3\\ 4x-2y=-6\end{array}\right.$

It can be easily seen that if we will substitute the value of $x$from the first equation to the second. We will get the value of $y$.

Therefore substitution will be the best method to solve part b.

It can be easily seen that if we will substitute the value of $x$from the first equation to the second. We will get the value of $y$.

Therefore substitution will be the best method to solve part $b$.