Chapter 7.6, Problem 55E

To determine

To solve: The given system of linear equations by substitution and check answer.

Answer to Problem 55E

  $x=0\&y=-7$

Explanation of Solution

Given information: Equations are $\begin{array}{l}5x-2y=14\\ -7=-2x+y\end{array}$ .

Calculation:

Revising first equation,

  $y=-\frac{14-5x}{2}$ $-(i)$

  $-7=-2x+y$ $-(ii)$

Substituting $(i)$ in $(ii)$ ,

  $-7=-2x+(-\frac{14-5x}{2})$

  $-7=\frac{-4x-14+5x}{2}$ (Taking LCM)

  $-14=-4x-14+5x$

  $x=0$

Putting above derived value of $x$ in $(i)$ ,

  $\begin{array}{l}y=-\frac{14-5(0)}{2}\\ y=\frac{-14}{2}\end{array}$

  $y=-7$

Verifying values, Putting $x=0$ and $y=-7$ in $(i)$

  $-7=-(\frac{14-5(0)}{2})$

  $-7=-7$

  $L.H.S.=R.H.S.$

Hence, true.

Putting $x=0$ and $y=-7$ in $(ii)$

  $\begin{array}{l}-7=-2(0)+(-7)\\ -7=-7\end{array}$

  $L.H.S.=R.H.S.$

Hence, true.

  $\therefore$ $x=0\&y=-7$