Chapter 6, Problem 12PT

To determine

Tosolve:The system of equations using elimination method.

Answer to Problem 12PT

The solution of given system of equations $x+y=8$ and $x-3y=-4$ is $x=5$ and $y=3$ .

Explanation of Solution

Given information:

The given system of equations is

  $x+y=8$

  $x-3y=-4$

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

  $x+y=8$  ...(1)

  $x-3y=-4$  ...(2)

As the coefficients of variable “ x ” in both equations are identical and of same signs so subtracting equations (2) from (1)

  $\left(x+y\right)-\left(x-3y\right)=8-\left(-4\right)$

  $\Rightarrow x-x+y+3y=8+4$

  $\Rightarrow \cancel{x}-\cancel{x}+y+3y=12$

  $\Rightarrow 4y=12$

  $\Rightarrow y=\frac{12}{4}=3$

Further, substituting value ofy in equation (1)

  $x+3=8$

  $\Rightarrow x=8-3=5$

Thus, $x=5$ , $y=3$ is the solution of system of equations $x+y=8$ and $x-3y=-4$ .

Conclusion:

The solution ofgiven system of equations $x+y=8$ and $x-3y=-4$ is $x=5$ and $y=3$ .