Chapter 3.2, Problem 11PPSE

Solution

To solve the system of equations $\left\{\begin{array}{c}3c+2d=6\\ d=4\end{array}\right.$ by the method of substitution and to check whether the answer is right.

The solution of the system of equations $\left\{\begin{array}{c}3c+2d=6\\ d=4\end{array}\right.$ is found and verified to be $c=-\frac{2}{3}$ and $d=4$ .

Given the system of equations:

  $\left\{\begin{array}{c}3c+2d=6\\ d=4\end{array}\right.$ .

Concept Used:

Method of substitution in the case of linear equations involving two variables:

Given a system of linear equations in two variables:

  $\left\{\begin{array}{c}{a}_{1}c+b_{1}d=c_1\\ a_{2}c+b_{2}d=c_2\end{array}\right.$ .

The method of substitution employs solving one of the equations, say $a_{1}c+b_{1}d=c_1$ , to find an algebraic expression for one of its variables, say $c$ .

Now, this expression of $c$ is substituted in the other equation $a_{2}c+b_{2}d=c_2$ and then solved to find the value of $d$ .

The value of $d$ is now substituted in either $a_{1}c+b_{1}d=c_1$ or $a_{2}c+b_{2}d=c_2$ to find the value of $c$ .

Now, the value of $c$ and that of $d$ are recorded to be the solution of the system of equations.

Calculation:

Solve the system of equations by the method of substitution:

Observe that the second equation, $d=4$ , is already suitable for being substituted into the first equation, $3c+2d=6$ .

Substitute $d=4$ in the first equation $3c+2d=6$ and solve for $c$ :

  $\begin{array}{c}3c+2d=6\\ 3c+2\cdot 4=6\\ 3c+8=6\\ 3c=6-8\\ 3c=-2\\ c=-\frac{2}{3}\end{array}$ .

Thus, it is found that:

  $c=-\frac{2}{3}$ .

Observe that:

  $d=4$ .

Thus, the solution of the system of equations $\left\{\begin{array}{c}3c+2d=6\\ d=4\end{array}\right.$ :

  $c=-\frac{2}{3}$ and $d=4$ .

Check whether the answer is correct or not:

Substitute $c=-\frac{2}{3}$ and $d=4$ in the first equation $3c+2d=6$ and determine whether it gives a valid equation:

  $\begin{array}{c}3c+2d=6\\ 3\cdot \left(-\frac{2}{3}\right)+2\cdot 4=6\\ -2+8=6\\ 6=6\end{array}$ .

It is observed that $3c+2d=6$ gives a valid equation when it is substituted with $c=-\frac{2}{3}$ and $d=4$ .

Thus, $c=-\frac{2}{3}$ and $d=4$ satisfy the equation $3c+2d=6$ .

Substitute $c=-\frac{2}{3}$ and $d=4$ in the second equation $d=4$ to check whether it gives a valid equation:

  $\begin{array}{l}d=4\\ 4=4\end{array}$ .

It is observed that $d=4$ gives a valid equation when it is substituted with $c=-\frac{2}{3}$ and $d=4$ .

Thus, $c=-\frac{2}{3}$ and $d=4$ satisfy the equation $d=4$ .

Hence, $c=-\frac{2}{3}$ and $d=4$ satisfy both the equations $3c+2d=6$ and $d=4$ .

That is, $c=-\frac{2}{3}$ and $d=4$ is a solution of the system of equations $\left\{\begin{array}{c}3c+2d=6\\ d=4\end{array}\right.$ .

Hence, the answer so found is correct.

Conclusion:

The solution of the system of equations $\left\{\begin{array}{c}3c+2d=6\\ d=4\end{array}\right.$ is found and verified to be $c=-\frac{2}{3}$ and $d=4$ .