Chapter 6, Problem 3CA

a.

To determine

To find: Two equations to support Student A’s claim.

Answer to Problem 3CA

  $2y-6x=12$

  $9x-3y=-18$

Explanation of Solution

Given information: Student A claims he can form a linear system from the equations shown that has infinitely many solutions. Student B claims she can form a linear system from the equations shown that has one solution. Student C claims he can form a linear system from the equations shown that has no solution.

  $3x+y=12$

  $3x+2y=12$

  $6x+2y=6$

  $3y+9x=36$

  $2y-6x=12$

  $9x-3y=-18$

Calculation: Considering the equations,

  $2y-6x=12$ ……………………….. (1)

  $9x-3y=-18$ ………………………. (2)

From equation (1),

  $\begin{array}{l}2y-6x=12\\ \Rightarrow 2y=12+6x\\ \Rightarrow y=6+3x\end{array}$

Putting the value of y in equation (2),

  $\begin{array}{l}9x-3y=-18\\ \Rightarrow 9x-3(6+3x)=-18\\ \Rightarrow 9x-18-9x=-18\\ \Rightarrow 9x+18=9x+18\end{array}$

This is true for all values of x. So the system has infinite solutions.

Thus, the system of linear equations which has infinite solution is,

  $2y-6x=12$

  $9x-3y=-18$

b.

To determine

To find: Two equations to support Student B’s claim.

Answer to Problem 3CA

  $3x+2y=12$

  $3y+9x=36$

Explanation of Solution

Given information: Student A claims he can form a linear system from the equations shown that has infinitely many solutions. Student B claims she can form a linear system from the equations shown that has one solution. Student C claims he can form a linear system from the equations shown that has no solution.

  $3x+y=12$

  $3x+2y=12$

  $6x+2y=6$

  $3y+9x=36$

  $2y-6x=12$

  $9x-3y=-18$

Calculation: Considering the equations,

  $3x+2y=12$ ……………………….. (3)

  $3y+9x=36$ ………………………. (4)

From equation (3),

  $\begin{array}{l}3x+2y=12\\ \Rightarrow 2y=12-3x\end{array}$

  $\Rightarrow y=6-\frac{3}{2}x$ ……………………. (5)

Putting the value of y in equation (4),

  $\begin{array}{l}3y+9x=36\\ \Rightarrow 3(6-\frac{3}{2}x)+9x=36\\ \Rightarrow 18-\frac{9}{2}x+9x=36\\ \Rightarrow \frac{9}{2}x=18\\ \Rightarrow x=18\times \frac{2}{9}\\ \Rightarrow x=4\end{array}$

Putting the value of x in equation (5),

  $\begin{array}{l}y=6-\frac{3}{2}x\\ \Rightarrow y=6-\frac{3}{2}\times 4\\ \Rightarrow y=6-6\\ \Rightarrow y=0\end{array}$

Thus, the solution of the system is $x=4,y=0$ .

This system has only one solution.

Thus, the system of linear equations which has one solution is,

  $3x+2y=12$

  $3y+9x=36$

c.

To determine

To find: Two equations to support Student C’s claim.

Answer to Problem 3CA

  $3x+y=12$

  $6x+2y=6$

Explanation of Solution

Given information: Student A claims he can form a linear system from the equations shown that has infinitely many solutions. Student B claims she can form a linear system from the equations shown that has one solution. Student C claims he can form a linear system from the equations shown that has no solution.

  $3x+y=12$

  $3x+2y=12$

  $6x+2y=6$

  $3y+9x=36$

  $2y-6x=12$

  $9x-3y=-18$

Calculation: Considering the equations,

  $3x+y=12$ ……………………….. (6)

  $6x+2y=6$ ………………………. (7)

From equation (6),

  $\begin{array}{l}3x+y=12\\ \Rightarrow y=12-3x\end{array}$

Putting the value of y in equation (7),

  $\begin{array}{l}6x+2y=6\\ \Rightarrow 6x+2(12-3x)=6\\ \Rightarrow 6x+24-6x=6\\ \Rightarrow 24=6\end{array}$

The statement is false . So the system has no solution.

Thus, the system of linear equations which has no solution is,

  $3x+y=12$

  $6x+2y=6$