Chapter 6, Problem 29SGR

To determine

To calculate : To solve the two equations using elimination method.

Answer to Problem 29SGR

The value of x is $\frac{1}{2}$ and the value of y is 6.

Explanation of Solution

Given information :

Given the two equations:

  $\begin{array}{l}6x+y=9\\ -6x+3y=15\end{array}$

Formula Used:

For elimination method, we multiply both the number with two different constant number the we add or subtract the equations, eliminate anyone first variable and then get the value of second variable. Further we put the value of the second variable in any of the equations and get the value of the first variable.

Calculation:

Equations:

  $6x+y=9$ .....Equation 1

  $-6x+3y=15$ .....Equation 2

Adding Equation 1 and Equation 2, we get:

  $\begin{array}{l}(6x+y)+(-6x+3y)=9+15\\ \Rightarrow 6x+y-6x+3y=24\\ \Rightarrow 4y=24\\ \Rightarrow y=\frac{24}{4}\\ \Rightarrow y=6\end{array}$

Putting the value of y into Equation 1, we get:

  $\begin{array}{l}6x+6=9\\ \Rightarrow 6x=9-6\\ \Rightarrow 6x=3\\ \Rightarrow x=\frac{3}{6}\\ \Rightarrow x=\frac{1}{2}\end{array}$

So the value of x is $\frac{1}{2}$ and the value of y is equal to 6.