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Solve the following system of equations.

$\left\{ \begin{aligned} {-3} & = {-m+{\frac{1}{2}}r}\\ {-m+{\frac{4}{3}}} & = {-r} \end{aligned} \right.$

Give your answer in the form m = ___ and r = ___ . If there are no solutions, you may enter no solutions or none. If there are infinitely many solutions, you may enter infinitely many solutions.

Solution

If an equation involves fractions, it is helpful to clear denominators by multiplying both sides of the equation by a common multiple of the denominators.

$\left\{\begin{aligned} 2\left(-3\right) & = 2\left(-m+{\frac{1}{2}}r\right)\\ 3\left(-m+{\frac{4}{3}}\right) & = 3\left(-r\right) \end{aligned}\right.$

$\left\{ \begin{aligned} -6 & = -2m+r\\ -3m+4 & = -3r \end{aligned} \right.$

Since one of the coefficients of $r$ is $1$ , it is wise to solve for $r$ in terms of the other variable and then use substitution to complete the problem.

$\begin{aligned} 2m-6 & = r &\text{(from the first equation)} \end{aligned}$

which we can substitute in for $r$ into the second equation:

$\begin{aligned} 4-3m & = -3\mathopen{}\left(2m-6\right) &\text{(from the second equation)}\\ 4-3m & = 18-6m\\3m & = 14\\ m &= {\frac{14}{3}} \end{aligned}$

We can substitute this back for $m$ into the first equation to find $r$ .

$\begin{aligned} 2\left({\frac{14}{3}}\right) - 6 & = r &\text{(from the first equation, after we had solved for }r\text{ in terms of }m\text{)}\\ {\frac{28}{3}}+\left(-6\right) &= r\\ {\frac{10}{3}} &= r \end{aligned}$

So the solution is $m={\frac{14}{3}}, r={\frac{10}{3}}$ .