Exercise Set 1.1 – No. 11

11. In each part, solve the linear system, if possible, and use the result to determine whether the lines represented by the equations in the system have zero, one, or infinitely many points of intersection. If there is a single point of intersection, give its coordinates, and if there are infinitely many, find parametric equations for them.

{\color{Blue} \begin{matrix} (a)\; 3x-2y = 4 \\ \, \, \, \, \, \, \, \, \, 6x -4y = 9 \end{matrix} }
 

Consider the linear system

{\color{Blue} \begin{matrix} 3x-2y = 4 \\ 6x -4y = 9 \end{matrix} } 

Suppose there exists (x, y) solution of the system above. Then we have that 3x – 2y = 4 and 6x-4y = 9 which gives us 2.(3x – 2y) – (6x – 4y) =2.4 – 9.

Since 2(3x – 2y) – (6x – 4y) = 6x – 4y – 6x + 4y = 0 and 2.4 – 9 = 8 – 9 =-1.

We have that 0 =  -1 which is a contradiction. Hence the system has no solution and therefore the lines whose equation are 3x – 2y = 4 and 6x – 4y = 9 have no intersection.

Here is sketch of the lines 3x – 2y = 4 and 6x – 4y = 9.

Exercise-Set-1.1-No.-11a.jpg
{\color{Blue} \begin{matrix} (b)\; 2x-4y = 1 \\ \, \, \, \, \, \, \, \, \, 4x -8y = 2 \end{matrix} }
 
 
 
 
 
 
 

Consider the linear system  {\color{Blue} \begin{matrix} 2x-4y = 1 \\ 4x -8y = 2 \end{matrix} }

We can eliminate x from the second equation by adding -2 times the first equation to second. This yields the simplified system  {\color{Blue} \begin{matrix} 2x-4y = 1 \\ \; \; \; \; \; \; \; \; \; \; 0 = 0 \end{matrix} }

The second equation does not any retrictions on x and y and hence can be omitted. Thus, the solutions of system are those values of x and y that satisfy the single equations 2x – 4y = 1. Geometrically, this means the lines correspoding to the two equations in the original system coincide which gives us that the system has infinitely many solutions.

Solving the equations 2x – 4y = 1 for x in terms of y gives us that x = 2x + 1/2. Assigning an arbitrary value t to y we can express the solutions fo the system in terms of the parametric equations x = 2t + 1/2 and y = t.

Hence (2t + 1/2, t) is a solution of the system above for all {\color{Blue} t \in \mathbb{R} }

Here’s a sketch of the line 2x – 4y = 1

Exercise Set 1.1 – No.11b

{\color{Blue} \begin{matrix} (c)\; x-2y = 0 \\ \, \, \, \, \, \, \, \, \, x -4y = 8 \end{matrix} }

Consider the linear system {\color{Blue} \begin{matrix} x-2y = 0 \\ x -4y = 8 \end{matrix} }

We can eliminate x from the second equation by adding -1 times the first equation to the second. This yields the simplified system {\color{Blue} \begin{matrix} x-2y = 0 \\ \; \; \; -2y = 8 \end{matrix} }

From the second equation obtain y = -4. Substituting y = -4 in the equation x – 2y = 0 gives us that x = -8.

Thus, the system has the unique solution x = -8 and y = -4.

Geometrically, this means that the lines represented by the equatiorvs in the system intersect at the single point (-8, -4).

Here is a sketch of the lines x – 2y = 0 and x – 4y = 8.

Exercise Set 1.1 – No.11c