Chapter 3.2, Problem 41PPS

To determine

To find: the numbers according to the given system of equation.

Answer to Problem 41PPS

  $x=\frac{46}{7};y=\frac{40}{7}$.

Explanation of Solution

Given:

Twice the sum of a number and three times a second number is 4.

And the difference of ten times the second number and five times the first number is 90.

Calculation:

Let the first number be x and second number be y .

According to the question,

  $\begin{array}{l}2x+3y=4\\ 10y-5x=90\end{array}$.

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

  $\begin{array}{l}5\left(2x+3y\right)=4\times 5\\ 10x+15y=20\\ 2\left(-5x+10y\right)=90\times 2\\ -10x+20y=180\end{array}$

Now, subtract the equation

  $\begin{array}{l}\underset{\_}{\begin{array}{l}10x+15y=20\\ -10x+20y=180\end{array}}\\ 35y=200\\ \therefore y=\frac{200}{35}\\ \Rightarrow y=\frac{40}{7}\end{array}$

Now, substitutes the value of y in 2nd equation, then:

  $\begin{array}{l}2x+3y=4\\ 2x+3\left(\frac{40}{7}\right)=4\{\because y=\frac{40}{7}\}\\ 2x+\frac{120}{7}=4\\ \therefore 2x=4-\frac{120}{7}=\frac{28-120}{7}=\frac{92}{7}\\ \therefore x=\frac{46}{7}\end{array}$

Hence, numbers according to the given system of equation is $x=\frac{46}{7};y=\frac{40}{7}$.