Chapter 9.5, Problem 1CE

To determine

To Verify: $\left(\frac{CE-BF}{AE-BD},\frac{AF-CD}{AE-BD}\right)$ is solution of given equations.

Answer to Problem 1CE

The equations $Ax+By=C$ and $Dx+Ey=F$ have solution $\left(\frac{CE-BF}{AE-BD},\frac{AF-CD}{AE-BD}\right)$ is verified.

Explanation of Solution

Given information:

The equations are

  $Ax+By=C$ and $Dx+Ey=F$ .

Calculation:

Consider the equations.

  $Ax+By=C$ ……. (1)

  $Dx+Ey=F$ …… (2)

Multiply equation (1) by $E$ and equation(2) by $B$

  $AEx+BEy=CE$ ……. (3)

  $BDx+BEy=BF$ …… (4)

Subtract (3) from (4).

  $\begin{array}{l}\left(BD-AE\right)x=BF-CE\\ x=\frac{BF-CE}{BD-AE}\\ x=\frac{CE-BF}{AE-BD}\end{array}$

Multiply equation (1) by $D$ and equation (2) by $A$

  $ADx+BDy=CD$ ……. (5)

  $ADx+AEy=AF$ …… (6)

Subtract (5) from (6).

  $\begin{array}{l}\left(BD-AE\right)y=CD-AF\\ y=\frac{CD-AF}{BD-AE}\\ x=\frac{AF-CD}{AE-BD}\end{array}$

Therefore, the equations $Ax+By=C$ and $Dx+Ey=F$ have solution $\left(\frac{CE-BF}{AE-BD},\frac{AF-CD}{AE-BD}\right)$ is verified.