Chapter 3.2, Problem 12PPSE

Solution

To solve the system of equations $\left\{\begin{array}{c}x+12y=68\\ x=8y-12\end{array}\right.$ by the method of substitution and to check whether the answer is right.

The solution of the system of equations $\left\{\begin{array}{c}x+12y=68\\ x=8y-12\end{array}\right.$ is found and verified to be $x=20$ and $y=4$ .

Given the system of equations:

  $\left\{\begin{array}{c}x+12y=68\\ x=8y-12\end{array}\right.$ .

Concept Used:

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

Given a system of linear equations in two variables:

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

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

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

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

Now, the value of $x$ and that of $y$ 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, $x=8y-12$ , is already suitable for being substituted into the first equation, $x+12y=68$ .

Substitute $x=8y-12$ in the first equation $x+12y=68$ and solve for $y$ :

  $\begin{array}{c}x+12y=68\\ \left(8y-12\right)+12y=68\\ 8y-12+12y=68\\ 8y+12y=68+12\\ 20y=80\\ y=\frac{80}{20}\\ y=4\end{array}$ .

Thus, it is found that:

  $y=4$ .

Substitute $y=4$ in the second equation $x=8y-12$ :

  $\begin{array}{c}x=8y-12\\ x=8\cdot 4-12\\ x=32-12\\ x=20\end{array}$ .

Thus, it is found that:

  $x=20$ .

Thus, the solution of the system of equations $\left\{\begin{array}{c}x+12y=68\\ x=8y-12\end{array}\right.$ :

  $x=20$ and $y=4$ .

Check whether the answer is correct or not:

Substitute $x=20$ and $y=4$ in the first equation $x+12y=68$ and determine whether it gives a valid equation:

  $\begin{array}{c}x+12y=68\\ 20+12\cdot 4=68\\ 20+48=68\\ 68=68\end{array}$ .

It is observed that $x+12y=68$ gives a valid equation when it is substituted with $x=20$ and $y=4$ .

Thus, $x=20$ and $y=4$ satisfy the equation $x+12y=68$ .

Substitute $x=20$ and $y=4$ in the second equation $x=8y-12$ to check whether it gives a valid equation:

  $\begin{array}{c}x=8y-12\\ 20=8\cdot 4-12\\ 20=32-12\\ 20=20\end{array}$ .

It is observed that $x=8y-12$ gives a valid equation when it is substituted with $x=20$ and $y=4$ .

Thus, $x=20$ and $y=4$ satisfy the equation $x=8y-12$ .

Hence, $x=20$ and $y=4$ satisfy both the equations $x+12y=68$ and $x=8y-12$ .

That is, $x=20$ and $y=4$ is a solution of the system of equations $\left\{\begin{array}{c}x+12y=68\\ x=8y-12\end{array}\right.$ .

Hence, the answer so found is correct.

Conclusion:

The solution of the system of equations $\left\{\begin{array}{c}x+12y=68\\ x=8y-12\end{array}\right.$ is found and verified to be $x=20$ and $y=4$ .