Chapter 2.2, Problem 3CFU

To determine

To describe: the steps to solve a system of three equations using elimination method.

Answer to Problem 3CFU

Firstly, make the coefficient of a variable in two equations look same; so that they can be we can use addition or subtraction to eliminated that variable. Then take third equation along with another and again try to make the coefficient of a same variable look same; so that we can eliminate that variable and make equations with two variables.

Explanation of Solution

Given information:

Step to solve system of three equation system.

Method used: Elimination

Calculation:

Firstly, make the coefficient of a variable in two equations look same; so that they can be we can use addition or subtraction to eliminated that variable. Then take third equation along with another and again try to make the coefficient of a same variable look same; so that we can eliminate that variable and make equations with two variables.

Let the function be $Ax+By+Cz=D$ , where $A,B,C\text{ and }D$ are constant. Now, try to eliminate one variable by multiplying the constants by its related constant $A_1\text{ and }{A}_2$ by its opposing equation as shown below.

  $\begin{array}{l}{C}_2\left(A_{1}x+B_{1}y+C_{1}z\right)=C_{2}D\mathrm{...}\left(1\right)\\ -C_1\left(A_{2}x+B_{2}y+C_{1}z\right)=-C_{1}D\mathrm{...}\left(2\right)\end{array}$

Now add equations (1) and (2), to get the third equation as follows:

  $\begin{array}{c}{C}_2{A}_{1}x-C_1{A}_{2}x+C_2{B}_{1}y-C_1{B}_{2}y+C_{1}z-C_{1}z=C_2{D}_1-C_1{D}_2\\ C_2{A}_{1}x-C_1{A}_{2}x+C_2{B}_{1}y-C_1{B}_{2}y=C_2{D}_1-C_1{D}_2\mathrm{...}\left(3\right)\end{array}$

Now add equations (2) and (3), to get the fourth equation as follows:

  $C_3{A}_{2}x-C_2{A}_{3}x+C_3{B}_{2}y-C_2{B}_{3}y=C_3{D}_2-C_2{D}_3\mathrm{...}\left(4\right)$

Equation (3) and (4) can be rewritten as follows:

  $\begin{array}{l}\left(C_2{A}_1-C_1{A}_2\right)x+\left(C_2{B}_1-C_1{B}_2\right)y=C_2{D}_1-C_1{D}_2\mathrm{...}\left(\text{Rewrting equation}\left(3\right)\right)\\ \left(C_3{A}_2-C_2{A}_3\right)x+\left(C_3{B}_2-C_2{B}_3\right)y=C_3{D}_2-C_2{D}_3\mathrm{...}\left(\text{Rewrting equation}\left(4\right)\right)\end{array}$

Now, solve both the equations for y, to get

  $\begin{array}{l}y=\frac{-\left(C_2{A}_1-C_1{A}_2\right)x+\left(C_2{D}_1-C_1{D}_2\right)}{\left(C_2{B}_1-C_1{B}_2\right)}\mathrm{...}\left(\text{Rewrting equation}\left(3\right)\right)\\ y=\frac{-\left(C_3{A}_2-C_2{A}_3\right)x+\left(C_3{D}_2-C_2{D}_3\right)}{\left(C_3{B}_2-C_2{B}_3\right)}\mathrm{...}\left(\text{Rewrting equation}\left(4\right)\right)\end{array}$

Equate both the equations to get the solution as follows:

  $\begin{array}{l}\frac{-\left(C_2{A}_1-C_1{A}_2\right)x+\left(C_2{D}_1-C_1{D}_2\right)}{\left(C_2{B}_1-C_1{B}_2\right)}=\frac{-\left(C_3{A}_2-C_2{A}_3\right)x+\left(C_3{D}_2-C_2{D}_3\right)}{\left(C_3{B}_2-C_2{B}_3\right)}\\ x=\frac{\left(C_2{B}_1-C_1{B}_2\right)\left(C_3{D}_2-C_2{D}_3\right)-\left(C_3{B}_2-C_2{B}_3\right)\left(C_2{D}_1-C_1{D}_2\right)}{-\left(C_2{A}_1-C_1{A}_2\right)\left(C_3{B}_2-C_2{B}_3\right)+\left(C_3{A}_2-C_2{A}_3\right)\left(C_2{B}_1-C_1{B}_2\right)}\end{array}$

NOTE: The above formula only works for $C_{number}\ne 0$