Chapter 3.2, Problem 35PPS

To determine

To find: the solution of the given system of equation using elimination.

Answer to Problem 35PPS

  $u=5;v=4$.

Explanation of Solution

Given:

The given system of equation is $\begin{array}{l}11u=5v+35\\ 8v=-6u+62\end{array}$.

Concept used:

Guidelines for solving system of equation using elimination:

1. First one of expression of the first equation is equivalent to the same expression of the second equation by multiplying.
2. Subtract both the equation.

Calculation:

As the given system of equation is expressed as $\begin{array}{l}11u-5v=35\\ 6u+8v=62\end{array}$.

Since none of the expressions is same on both the equation, therefore multiply by 8 in 1st equation and multiply by 5 in 2nd equation and then equate:

  $\begin{array}{l}8\left(11u-5v\right)=35\times 8\\ 88u-40v=280\\ 5\left(6u+8v\right)=62\times 5\\ 30u+40v=310\end{array}$

Now, add the equation

  $\begin{array}{l}\underset{\_}{\begin{array}{l}88u-40v=280\\ 30u+40v=310\\ +++\end{array}}\\ 118u=590\\ \therefore u=\frac{590}{118}\\ \Rightarrow u=5\end{array}$

Now, substitutes the value of u in 2^ndequation, then:

  $\begin{array}{l}8v=-6u+62\\ 8v=-6\left(5\right)+62\{\because u=5\}\\ 8v=-30+62\\ \therefore 8v=32\\ \therefore v=\frac{32}{4}=8\end{array}$

Hence, the value of given system of equation is $u=5;v=4$.