Chapter 7.1, Problem 73E

(a)

To determine

To calculate: Under what condition on the coefficients of the system , a, b. c, d, e, and f the above system of linear equations has a one distinct solution.

Answer to Problem 73E

Condition to have a distinct solution of the above system of linear equations is $ae-bd\ne 0$ .

Explanation of Solution

Given information:

In the question following general form of the system of linear equations is given

  $\begin{array}{l}ax+by=c\\ dx+ey=f\end{array}$

Calculation:

Given system of equations is

  $\begin{array}{l}ax+by=c\\ dx+ey=f\end{array}$

has a distinct solution if $\frac{a}{d}\ne \frac{b}{e}$ , multiplying both sides by $de$ , that is $de\frac{a}{d}\ne de\frac{b}{e}$ , it will finally give the condition $ae\ne db$ .

(b)

To determine

To calculate: the solution of the given system of the linear equations by method of substitution and the graphical method.

Answer to Problem 73E

The solution of the given system of equations is

  $\begin{array}{l}x=\frac{ce-bf}{ae-bd}\\ y=\frac{af-cd}{ae-bd}\end{array}$

Explanation of Solution

1. Method of Substitution

Given system of linear equations is

  $\begin{array}{l}ax+by=c\dots (1)\\ dx+ey=f\dots (2)\end{array}$

To solve the system of equation by the method of substitution take the following steps:

Step-1: Solve for y from the equation (1):

  $ax+by=c$

On subtracting ax from both sides of the above equation, it will give

  $\begin{array}{l}ax+by-ax=c-ax\\ (ax-ax)+by=c-ax\\ 0+by=c-ax\\ by=c-ax\end{array}$

Now, divide both sides by b to get the value of y :

  $\begin{array}{l}\frac{by}{b}=\frac{c-ax}{b}\\ y=\frac{c-ax}{b}\end{array}$

Step-2: Substitute the above value of y in the second equation (2), and solve it for the value of x :

  $\begin{array}{l}dx+ey=f\\ dx+e\left(\frac{c-ax}{b}\right)=f\\ dx+\frac{e\left(c-ax\right)}{b}=f\end{array}$

On multiplying both sides by b, it will give

  $\begin{array}{l}b\left(dx+\frac{e(c-ax)}{b}\right)=bf\\ bdx+\frac{be(c-ax)}{b}=bf\\ bdx+e(c-ax)=bf\\ bdx+ec-eax=bf\\ (bd-ea)x+ec=bf\end{array}$

Now, subtracting ec from both sides:

  $\begin{array}{l}(bd-ae)x+ec-ec=bf-ec\\ (bd-ae)x+0=bf-ec\\ (bd-ae)x=bf-ec\end{array}$

Now, on dividing both sides by (bd-ae), it will the value of x :

  $\begin{array}{l}\frac{(bd-ae)x}{bd-ae}=\frac{bf-ec}{bd-ae}\\ x=\frac{bf-ec}{bd-ae}=\frac{ec-bf}{ae-bd}\end{array}$

Or,

  $x=\frac{ec-bf}{ae-bd}$

Notice that above value of x is meaningful only when $ae-bd\ne 0$ .

Step-3: (Back-substitution) Substitute the value of x in the value of y calculated in the step-1:

  $\begin{array}{l}y=\frac{(c-ax)}{b}=\frac{\left(c-a\left(\frac{ce-bf}{ae-bd}\right)\right)}{b}\\ y=\frac{\left(c-\frac{a(ce-bf)}{ae-bd}\right)}{b}=\frac{\left(\frac{c(ae-bd)-a(ce-bf)}{ae-bd}\right)}{b}\\ y=\frac{\left(\frac{cae-cbd-ace+abf}{ae-bd}\right)}{b}=\frac{\left(\frac{cae-ace+abf-cbd}{ae-bd}\right)}{b}\\ y=\frac{\left(\frac{0+abf-cbd}{ae-bd}\right)}{b}=\frac{\left(\frac{abf-cbd}{ae-bd}\right)}{b}=\frac{\left(\frac{b(af-cd)}{ae-bd}\right)}{b}\\ y=\frac{\frac{b(af-cd)}{ae-bd}}{b}=\frac{b(af-cd)}{b(ae-bd)}=\frac{af-cd}{ae-bd}\end{array}$

Or,

  $y=\frac{af-cd}{ae-bd}$

It gives the values of x and y by the method of substitution.

1. Graphical Method:

To solve the given system of equation using graphical method, write the given linear equations in the following forms:

Equation (1)

  $\begin{array}{l}ax+by=c\\ by=-ax+c\\ y=-\left(\frac{a}{b}\right)x+\frac{c}{b}\dots (3)\end{array}$

And equation (2)

  $\begin{array}{l}dx+ey=f\\ ey=-dx+f\\ y=-\left(\frac{d}{e}\right)x+\frac{f}{e}\dots (4)\end{array}$

Notice that equation (3) and (4) are the equations of the straight lines. The system of linear equations (eqn (1)-(2)) has a solution if the straight lines (3) and (4) intersects. If the lines (3) and (4) intersects then the left hand sides of (3) and (4) are equal:

  $-\left(\frac{a}{b}\right)x+\frac{c}{b}=-\left(\frac{d}{e}\right)x+\frac{f}{e}$

Now, solving the above equation for x ,

  $\begin{array}{l}-\left(\frac{a}{b}\right)x+\frac{c}{b}=-\left(\frac{d}{e}\right)x+\frac{f}{e}\\ -\left(\frac{a}{b}\right)x+\left(\frac{d}{e}\right)x=\frac{f}{e}-\frac{c}{b}\\ \left(\frac{d}{e}-\frac{a}{b}\right)x=\frac{f}{e}-\frac{c}{b}\\ x=\frac{\frac{f}{e}-\frac{c}{b}}{\frac{d}{e}-\frac{a}{b}}=\frac{\frac{c}{b}-\frac{f}{e}}{\frac{a}{b}-\frac{d}{e}}\end{array}$

Or,

  $x=\frac{\frac{c}{b}-\frac{f}{e}}{\frac{a}{b}-\frac{d}{e}}\dots (5)$

And, the corresponding y-coordinate of the intersection point can be calculate from the equation (3):

  $\begin{array}{l}y=-\left(\frac{d}{e}\right)x+\frac{f}{e}\\ y=-\left(\frac{d}{e}\right)\left(\frac{\frac{c}{b}-\frac{f}{e}}{\frac{a}{b}-\frac{d}{e}}\right)+\frac{f}{e}\\ y=\frac{-\left(\frac{d}{e}\right)\left(\frac{c}{b}-\frac{f}{e}\right)+\left(\frac{f}{e}\right)\left(\frac{a}{b}-\frac{d}{e}\right)}{\frac{a}{b}-\frac{d}{e}}\\ y=\frac{-\frac{d}{e}\frac{c}{b}+\frac{d}{e}\frac{f}{e}+\frac{f}{e}\frac{a}{b}-\frac{f}{e}\frac{d}{e}}{\frac{a}{b}-\frac{d}{e}}\\ y=\frac{\frac{d}{e}\frac{f}{e}-\frac{f}{e}\frac{d}{e}+\frac{f}{e}\frac{a}{b}-\frac{d}{e}\frac{c}{b}}{\frac{a}{b}-\frac{d}{e}}\\ y=\frac{0+\frac{f}{e}\frac{a}{b}-\frac{d}{e}\frac{c}{b}}{\frac{a}{b}-\frac{d}{e}}\\ y=\frac{\frac{f}{e}\frac{a}{b}-\frac{d}{e}\frac{c}{b}}{\frac{a}{b}-\frac{d}{e}}\dots (6)\end{array}$

Notice that the point of intersection is equal to the solution of system of linear equations (1)-(2) determined by the method of substitutions. Point of intersection ( x,y ) in the following figure is given by the equations (5) and (6):

  

(c)

To determine

To write down the advantages of method of substitution compared to the graphical method.

Explanation of Solution

With graphical method, it is possible to locate or represent the solution as a point of intersection, but to actually find the solution one has to solve the equations using method of substitution. Sometimes in case of extremely large or small solutions it is not possible to draw a clear graph to represent the solution of the system of linear equations as the point of intersection. On the other hand, if the solution exists then it is always possible to find the solution using the method of substitution with a few calculation steps irrespective the size of solution.