Chapter 3.1, Problem 67PPS

To determine

To solve: The given system of equations

Answer to Problem 67PPS

The solution of the given system is given by,

  $\left(16,-8\right)$

Explanation of Solution

Given Information:

The system of equations

  $\begin{array}{l}\frac{5}{2}a-\frac{3}{4}b=46\\ -\frac{7}{8}a-3b=10\end{array}$

Formula Used

Here, two equations with two unknown variables are given. The approach used is variable elimination and simplification.

(1) Both equations are multiplied with a factor such that one of the variables has identical coefficients.

(2) Add the modified equations to eliminate the particular variable and solve for the other variable.

(3) Substitute the obtained variable in one of the equations to get the other variable.

Calculation Here, the equations are,

  $\frac{5}{2}a-\frac{3}{4}b=46$........(1)

  $-\frac{7}{8}a-3b=10$........(2)

Multiplying each of the terms in equation (2) by- $\frac{1}{4}$ , it becomes

  $-\frac{7}{8}\times \left(-\frac{1}{4}\right)-\left(-\frac{1}{4}\right)\times 3b=10\times \left(-\frac{1}{4}\right)$

  $\frac{7}{32}a+\frac{3}{4}b=-\frac{5}{2}$........(3)

Now, adding corresponding terms of (3) and (1) gives,

  $\{\begin{array}{l}\frac{5}{2}a-\frac{3}{4}b=46\\ \frac{7}{32}a+\frac{3}{4}b=-\frac{5}{2}\end{array}$

  $\begin{array}{l}\frac{7}{32}a+\frac{5}{2}a+0=-\frac{5}{2}+46\\ \frac{7+80}{32}a=\frac{87}{2}\\ a=16\end{array}$

Substituting the value of $a=16$ in(2), the value of b can be obtained as,

  $\begin{array}{c}-\frac{7}{8}\left(16\right)-3b=10\\ b=-\frac{10+14}{3}\\ =-8\end{array}$

Thus, the solution for the given system of equations can be expressed as,

  $\left(16,-8\right)$