Chapter EP, Problem 3.2.9EP

To determine

Solve equation using substitution method or elimination method.

Answer to Problem 3.2.9EP

  $\boxed{\begin{array}{l}c=\frac{1}{3}\\ d=-\frac{1}{2}\end{array}}$

Explanation of Solution

Given information:

The given equations are:

  $\begin{array}{l}3c+4d=-1\\ 6c-2d=3\end{array}$

  $3c+4d=-1$ ….1

  $6c-2d=3$ ….2

As we know in substitution method, we have to find value of one variable from one of the equations in form of other and substitute that value in another equation to find value of both the variable.

Now from the above equation we will find the value of $c$ in terms of $d$ from equation 1:

  $\begin{array}{l}3c+4d=-1\\ c=\frac{-1-4d}{3}\end{array}$ ....3

Putting value of $c$ from equation 3 in equation 2:

  $\begin{array}{l}6c-2d=3\\ 6\left(\frac{-1-4d}{3}\right)-2d=3\\ \frac{-6-24d}{3}-2d=3\\ \frac{-6-24d-6d}{3}=3\\ -6-30d=9\\ -30d=9+6\\ -30d=15\\ d=-\frac{15}{30}\\ \boxed{d=-\frac{1}{2}}\end{array}$

Now putting value of $d$ in equation 1 to find value of $c$ :

  $\begin{array}{l}3c+4d=-1\\ 3c+4\left(-\frac{1}{2}\right)=-1\\ 3c-2=-1\\ 3c=2-1\\ 3c=1\\ \boxed{c=\frac{1}{3}}\end{array}$