Chapter 6, Problem 10PT

To determine

Tosolve:The system of equations using elimination method.

Answer to Problem 10PT

The solution of system of equations $x+y=13$ and $x-y=5$ is $x=9$ and $y=4$ .

Explanation of Solution

Given information:

The given system of equations is

  $x+y=13$

  $x-y=5$

Methodused:

The steps of performing elimination method are:

Step 1: Write each given equation in standard form $Ax+By=C$ and $A^{\prime }x+B^{\prime }y=C^{\prime }$ .

Step 2:Equate coefficients of one variable equal by multiplying coefficients of variables in both equations. Let for variable “ x ”

  $A^{\prime }Ax+A^{\prime }By=A^{\prime }C$ , and $A{A}^{\prime }x+A{B}^{\prime }y=A{C}^{\prime }$ , then subtracting one equation from other variable “ x ” gets eliminated and remaining equation can be solved for “ y ”.

Calculations:

The given equations are

  $x+y=13$  ...(1)

  $x-y=5$  ...(2)

As the coefficients of variable “ y ” in both equations are identical and of opposite sign so adding equations (1) and (2)

  $\left(x+y\right)+\left(x-y\right)=13+5$

  $\Rightarrow \left(x+x\right)+\left(y-y\right)=18$

  $\Rightarrow 2x+0=18$

  $x=\frac{18}{2}=9$

Further, substituting value ofx in equation (1)

  $9+y=13$

  $\Rightarrow y=13-9=4$

Thus, $x=9$ , $y=13$ is the solution of system of equations $x+y=13$ and $x-y=5$ .

Conclusion:

The solution ofsystem of equations $x+y=13$ and $x-y=5$ is $x=9$ and $y=4$ .