91Ó°ÊÓ

The algebraic method of elimination involves adding or subtracting the equations to eliminate one of the variables and forming new equation that is true. Sometimes, direct addition or subtraction of equations does not eliminate the variable then one equation requires formation of equivalent equation through multiplication so that one of the two variables has the same or opposite coefficient in both the equations. Multiplying the equation by a nonzero number, resulting new equation has same set of solutions.

To solve the equations, multiply $\frac{1}{6}f+\frac{1}{3}g=1$by $6$then subtract the resulting equation from the first equation as shown below.

$\begin{array}{c}6\left(\frac{1}{6}f+\frac{1}{3}g\right)=6\left(1\right)\\ f+2g=6\end{array}$

Now, subtract $f=6-2g$and $f+2g=6$.

$\begin{array}{ccccccc}& f& +& 2g& =& & 6\\ & f& +& 2g& =& & 6\\ −& & −& & & −& \\ & 0& +& 0& =& & 0\end{array}$

Simplify it further as

$0=0$

Since, the equation$0=0$ is true, so the given system of equations is a dependent system and infinitely many solutions exist as both the equations are equivalent to each other and lies on the shared(same) axis.

A dependent system of equations is one in which the result on solving the equations, is an equation that is always true, therefore infinitely many solutions exist for such system of equations.

Hence, there are infinitely many solutions for the provided system of equations.