Chapter 3.1, Problem 48PPSE

Solution

To write the second equation of the system of equations, $\left\{\begin{array}{c}\frac{x}{4}+\frac{y}{3}=1\\ -----\end{array}\right.$ , so that it will have infinitely many solutions.

The choice $3x+4y=12$ for the second equation of the system of linear equations $\left\{\begin{array}{c}\frac{x}{4}+\frac{y}{3}=1\\ -----\end{array}\right.$ yields infinitely many solutions for it.

Given the incomplete system of linear equations:

  $\left\{\begin{array}{c}\frac{x}{4}+\frac{y}{3}=1\\ -----\end{array}\right.$ .

Concept Used:

System of linear equations having infinitely many solutions in the case of two variables:

Given a system of linear equations in two variables:

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

Then, the system of linear equations has infinitely many solutions if and only if one of the equations, say $a_{2}x+b_{2}y=c_2$ , is a scalar multiple of the other, $a_{1}x+b_{1}y=c_1$ .

Calculation:

Consider the incomplete system of equations:

  $\left\{\begin{array}{c}\frac{x}{4}+\frac{y}{3}=1\\ -----\end{array}\right.$ .

Multiply the first equation $\frac{x}{4}+\frac{y}{3}=1$ with a scalar, say $12$ :

  $12\left(\frac{x}{4}+\frac{y}{3}\right)=12\left(1\right)$

Simplify the equation $12\left(\frac{x}{4}+\frac{y}{3}\right)=12\left(1\right)$ :

  $\begin{array}{c}12\left(\frac{x}{4}+\frac{y}{3}\right)=12\left(1\right)\\ 12\left(\frac{x}{4}\right)+12\left(\frac{y}{3}\right)=12\left(1\right)\\ 3x+4y=12\end{array}$

Choose $3x+4y=12$ to be the second equation of the system of linear equations.

Then, the system of equations will have infinitely many solutions.

Conclusion:

The choice $3x+4y=12$ for the second equation of the system of linear equations $\left\{\begin{array}{c}\frac{x}{4}+\frac{y}{3}=1\\ -----\end{array}\right.$ yields infinitely many solutions for it.