Chapter 6.3, Problem 4P

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 $4x+3y=-19,3x-2y=-10$

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 $4x+3y=-19$

Multiply both sides by $2$

  $\begin{array}{l}2\left(4x+3y\right)=2\left(-19\right)\\ \Rightarrow 8x+6y=-38(i)\end{array}$

Consider the equation $3x-2y=-10$

Multiply both sides by $3$

  $\begin{array}{l}3\left(3x-2y\right)=3\left(-10\right)\\ \Rightarrow 9x-6y=-30(ii)\end{array}$

b.

To determine

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

Answer to Problem 4P

  $x=-4,y=-1$

Explanation of Solution

Calculation:

From $(i)$ and $(ii)$ ., the equations are $8x+6y=-38,9x-6y=-30$

Add both the equations.

  $\begin{array}{l}\left(8x+6y\right)+\left(9x-6y\right)=-38-30\\ \Rightarrow 17x=-68\\ \Rightarrow x=\frac{-68}{17}=-4\end{array}$

Put $x=-4$ in $8x+6y=-38$

  $\begin{array}{l}8\left(-4\right)+6y=-38\\ \Rightarrow -32+6y=-38\\ \Rightarrow 6y=-38+32=-6\\ \Rightarrow y=\frac{-6}{6}=-1\end{array}$

c.

To determine

To verify: $x=-4,y=-1$ is a solution of the original system of equation $4x+3y=-19,3x-2y=-10$

Explanation of Solution

Proof:

Put $x=-4,y=-1$ in $4x+3y=-19$

  $\begin{array}{l}4x+3y=-19\\ \Rightarrow 4\left(-4\right)+3\left(-1\right)=-19\\ \Rightarrow -16-3=-19\\ \Rightarrow -19=-19\text{True}\end{array}$

So, $x=-4,y=-1$ is a solution of $4x+3y=-19$

Now, put $x=-4,y=-1$ in $3x-2y=-10$

  $\begin{array}{l}3\left(-4\right)-2\left(-1\right)=-10\\ \Rightarrow -12+2=-10\\ \Rightarrow -10=-10\text{True}\end{array}$

So, $x=-4,y=-1$ is a solution of the equation $3x-2y=-10$