Chapter 5.3, Problem 16E

To determine

To Solve: The given system of linear equations by elimination.

  $\begin{array}{l}-2x-5y=9\\ 3x+11y=4\end{array}$

Answer to Problem 16E

The required solution is $x=-17,y=5$

Explanation of Solution

Given information:

  $\begin{array}{l}-2x-5y=9\\ 3x+11y=4\end{array}$

Calculation:

  $\begin{array}{l}-2x-5y=9\text{ Equation 1}\\ 3x+11y=4\text{ Equation 2 }\end{array}$

Multiply Equation 1 by 3 and Equation 2 by 2 so that the co-efficient of x-terms are opposites

  $\begin{array}{l}-2x-5y=9\\ \Rightarrow 3\left(-2x-5y\right)=3\cdot 9\\ \Rightarrow -6x-15y=27\text{ Revised Equation 1}\\ \\ 3x+11y=4\\ \Rightarrow 2\left(3x+11y\right)=4\cdot 2\\ \Rightarrow 6x+22y=8\text{ Revised Equation 2}\end{array}$

Variable x can be eliminated by adding equation 1 and revised equation 2

  $\begin{array}{l}-6x-15y=27\\ \underset{\_}{6x+22y=8}\\ \text{ 7}y=35\end{array}$

Divide 7 on both sides:

  $\begin{array}{l}\Rightarrow \frac{7y}{7}=\frac{35}{7}\\ \Rightarrow y=5\end{array}$

Now substitute the value of y in equation 1

  $\begin{array}{l}-2x-5y=9\\ \Rightarrow -2x-5\cdot 5=9\\ \Rightarrow -2x-25=9\end{array}$

Add 25on both sides

  $\begin{array}{l}\Rightarrow -2x-25+25=9+25\\ \Rightarrow -2x=34\end{array}$

Divide -2 on both sides

  $\begin{array}{l}\Rightarrow \frac{-2x}{-2}=\frac{34}{-2}\\ \Rightarrow x=-17\end{array}$

Check

Right hand side of equation 1 is 9

Left hand side of equation 1 is $-2x-5y$

Substituting the known value and simplify

  $\begin{array}{l}=-2\cdot \left(-17\right)-5\cdot \left(5\right)\\ =34-25\\ =9\end{array}$

Left hand side = Right hand side.

Right hand side of equation 2 is 4

Left hand side of equation 2 is $3x+11y$

Substituting the known value and simplify

  $\begin{array}{l}=3\cdot \left(-17\right)+11\cdot \left(5\right)\\ =-51+55\\ =4\end{array}$

Left hand side = Right hand side.