[tex] \huge \boxed{\mathbb{QUESTION} \downarrow}[/tex]
[tex] \large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}[/tex]
We can use the substitution method to solve linear equations of this form. Let's solve for m & n.
[tex]\left. \begin{array} { l } { 2 m + n = 2 } \\ { 3 m - 2 n = 3 } \end{array} \right.[/tex]
To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
[tex]2m+n=2, \: 3m-2n=3 [/tex]
Choose one of the equations and solve it for m by isolating m on the left-hand side of the equal sign.
[tex]2m+n=2 [/tex]
Subtract n from both sides of the equation.
[tex]2m=-n+2 [/tex]
Divide both sides by 2.
[tex]m=\frac{1}{2}\left(-n+2\right) \\ [/tex]
Multiply 1/2 times -n+2.
[tex]m=-\frac{1}{2}n+1 \\ [/tex]
Substitute [tex]-\frac{n}{2}+1\\[/tex] for m in the other equation, 3m-2n=3.
[tex]3\left(-\frac{1}{2}n+1\right)-2n=3 \\ [/tex]
Multiply 3 times [tex]-\frac{n}{2}+1\\[/tex].
[tex]-\frac{3}{2}n+3-2n=3 \\ [/tex]
Add [tex]-\frac{3n}{2}\\[/tex] to -2n.
[tex]-\frac{7}{2}n+3=3 \\ [/tex]
Subtract 3 from both sides of the equation.
[tex]-\frac{7}{2}n=0 \\ [/tex]
Divide both sides of the equation by [tex]-\frac{7}{2}\\[/tex], which is the same as multiplying both sides by the reciprocal of the fraction.
[tex] \large \underline{\underline{ \bf \: n=0 }}[/tex]
Substitute 0 for n in [tex]m=-\frac{1}{2}n+1\\[/tex]. Because the resulting equation contains only one variable, you can solve for m directly.
[tex] \large \underline{ \underline{\bf \: m=1 }}[/tex]
The system is now solved.
[tex] \huge \boxed{ \boxed{ \bf \: m=1, \: n=0 }}[/tex]
