Chapter 6.2, Problem 38PPE

a)

To determine

To find:The situation in which a graph depicts that a system of linear equations has no solution.

Answer to Problem 38PPE

If the lines representing the given system of linear equations are parallel shown in graph, such system has NO SOLUTUION as both lines have no intersection point as their unique solution.

Explanation of Solution

Giveninformation: Graph of any system of twolinear equations is given..

Concept used:Solution of a system of linear equations in two variables graphically is the intersecting point of their corresponding lines. But if these lines are parallel then they will never intersect each other at any of the point and thus there will be no intersecting point and hence there will be NO solutions to such system of linear equations.

For being NO solution of a given system of linear equations in two variables, its graph should show two parallel lines as they never intersect.

Conclusion: Using graph, if graph shows two parallel lines, system of linear equations has NO SOLUTION.

b)

To determine

To find:The situation when a system of linear equations has no solution using substitution.

Answer to Problem 38PPE

If the ratios of coefficients of x terms and y terms are equal but not equal to constant terms, then such system of linear equations has NO SOLUTION.

Explanation of Solution

Giveninformation: Any system of linear equations in two variables is given.

Concept/calculationused:

If $a_{1}x+b_{1}y=c_1\text{and }{a}_{2}x+b_{2}y=c_2$ are two linear equations. By first equation,

  $x=\frac{c_1-b_{1}y}{a_1}$ . Substitute this value of x in second equation and solve for y as:

  $\begin{array}{l}{a}_2\left(\frac{c_1-b_{1}y}{a_1}\right)+b_{2}y=c_2\\ \frac{a_2{c}_1}{a_1}-\frac{a_2{b}_{1}y}{a_1}+b_{2}y=c_2\\ -\frac{a_2{b}_{1}y}{a_1}+b_{2}y=c_2-\frac{a_2{c}_1}{a_1}\\ \Rightarrow \frac{a_1{b}_{2}y-a_2{b}_{1}y}{a_1}=\frac{a_1{c}_2-a_2{c}_1}{a_1}\end{array}$

  $\begin{array}{l}\Rightarrow (a_2{b}_2-a_2{b}_1)y=a_1{c}_2-a_2{c}_1\\ y=\frac{a_1{c}_2-a_2{c}_1}{a_1{b}_2-a_2{b}_1},a_1{b}_2-a_2{b}_1\ne 0\end{array}$

Conclusion: Using above condition, that is calculated using substitution method, it can be concluded that system of linear equations has NO SOLUTION only if $a_1{b}_2-a_2{b}_1\ne 0$ .

c)

To determine

To find:If a system of linear equations has one intersecting point, no point of infinite number of intersecting points.

Answer to Problem 38PPE

Explanation of Solution

Giveninformation: Two tables showing values of its variables of two linear equations.

Concept used:If $a_{1}x+b_{1}y=c_1\text{and }{a}_{2}x+b_{2}y=c_2$ are two linear equations, then if such system has NO SOLUTION, then condition using substitution is:

  $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

If system has one intersecting point then,

  $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

And if system has infinite intersecting points, then,

  $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

If one ordered pair in both tables of values is common, it means that two linear intersect at a single point. If the corresponding x values and y values of different ordered pairs of both tables are having equal ratios, such system has infinite number of intersecting points. And by observing the values of two tables, it cannot be decided if system has no solution as it does not depend on variables values.