Chapter 3.2, Problem 67HP

To determine

To write: an equation in which one is multiply by 3 and other is multiply by 4. Also, find its solution.

Answer to Problem 67HP

  $a=-4;b=-3$.

Explanation of Solution

Given:

Let the system of equation is $\begin{array}{l}3a+5b=-27\\ 4a+10b=-46\end{array}$.

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 or add both the equation.

Calculation:

As the given system of equation is $\begin{array}{l}3a+5b=-27\\ 4a+10b=-46\end{array}$.

Since none of the expressions is same on both the equation, therefore multiply by 4 in 1st equation and multiply by 3 in 2nd equation then equate:

  $\begin{array}{l}4\left(3a+5b\right)=-27\times 4\\ 12a+20b=-108\\ 3\left(4a+10b\right)=-46\times 3\\ 12a+30b=-138\end{array}$

  $\begin{array}{l}\underset{\_}{\begin{array}{l}12a+20b=-108\\ 12a+30b=-138\\ --+\end{array}}\\ -10b=30\\ \therefore b=\frac{30}{-10}\\ \Rightarrow b=-3\end{array}$

Now, substitutes the value of z in 1st equation, then:

  $\begin{array}{l}3a+5b=-27\\ 3a+5\left(-3\right)=-27\{\because b=-3\}\\ 3a-15=-27\\ 3a=-12\\ \therefore a=-4\end{array}$

Hence, the value of given system of equation is $a=-4;b=-3$.