Chapter 6.3, Problem 3P

a.

To determine

To calculate: changed form of an equation in the given system of equations to solve it using elimination

Explanation of Solution

Given information: Equations are $-5x-2y=6,3x+6y=6$

Calculation:

According to multiplication property of equality, if we multiply the equation by a non-zero number, we get an equivalent equation.

Consider the equation $-5x-2y=6$

Multiply both sides by $3$

  $\begin{array}{l}3\left(-5x-2y\right)=3\left(6\right)\\ \Rightarrow -15x-6y=18\end{array}$

b.

To determine

To calculate: value of $x$ and $y$

Answer to Problem 3P

  $x=-2,y=2$

Explanation of Solution

Calculation:

From part a., the equation $-5x-2y=6$ becomes $-15x-6y=18$

The other equation is $3x+6y=6$

Add both the equations

  $\begin{array}{l}\left(-15x-6y\right)+\left(3x+6y\right)=18+6\\ \Rightarrow -15x-6y+3x+6y=24\\ \Rightarrow -12x=24\\ \Rightarrow x=\frac{24}{-12}=-2\end{array}$

Put $x=-2$ in the equation $-5x-2y=6$

  $\begin{array}{l}-5\left(-2\right)-2y=6\\ \Rightarrow 10-2y=6\\ \Rightarrow 2y=10-6\\ \Rightarrow 2y=4\\ \Rightarrow y=\frac{4}{2}=2\end{array}$

c.

To determine

To verify: $x=-2,y=2$ is a solution of the original system of equations $-5x-2y=6,3x+6y=6$

Explanation of Solution

Proof:

Put $x=-2,y=2$ in $-5x-2y=6$

  $\begin{array}{l}-5\left(-2\right)-2\left(2\right)=6\\ \Rightarrow 10-4=6\\ \Rightarrow 6=6\text{True}\end{array}$

So, $x=-2,y=2$ is a solution of $-5x-2y=6$

Now, put $x=-2,y=2$ in $3x+6y=6$

  $\begin{array}{l}3\left(-2\right)+6\left(2\right)=6\\ \Rightarrow -6+12=6\\ \Rightarrow 6=6\text{True}\end{array}$

So, $x=-2,y=2$ is a solution of the equation $3x+6y=6$