To determine the values of [tex]\( m \)[/tex] and [tex]\( n \)[/tex] that make the pair of linear equations have an infinite number of solutions, we need to ensure they are equivalent equations. For this to happen, the equations must have the same ratio for the coefficients of [tex]\( x \)[/tex], [tex]\( y \)[/tex], and their constant terms.
The given equations are:
1. [tex]\((2m - 1)x + 3y = 5\)[/tex]
2. [tex]\(3x + (n - 1)y = 2\)[/tex]
Let's denote the coefficients of the first equation as [tex]\(a_1\)[/tex], [tex]\(b_1\)[/tex], and [tex]\(c_1\)[/tex], and for the second equation as [tex]\(a_2\)[/tex], [tex]\(b_2\)[/tex], and [tex]\(c_2\)[/tex]. Therefore:
- [tex]\(a_1 = 2m - 1\)[/tex]
- [tex]\(b_1 = 3\)[/tex]
- [tex]\(c_1 = 5\)[/tex]
- [tex]\(a_2 = 3\)[/tex]
- [tex]\(b_2 = n - 1\)[/tex]
- [tex]\(c_2 = 2\)[/tex]
For the equations to have an infinite number of solutions, the following ratios must hold true:
[tex]\[
\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}
\][/tex]
Using the ratios:
1. [tex]\[\frac{a_1}{a_2} = \frac{2m - 1}{3}\][/tex]
2. [tex]\[\frac{b_1}{b_2} = \frac{3}{n - 1}\][/tex]
3. [tex]\[\frac{c_1}{c_2} = \frac{5}{2}\][/tex]
First, let's equate [tex]\(\frac{a_1}{a_2}\)[/tex] and [tex]\(\frac{c_1}{c_2}\)[/tex]:
[tex]\[
\frac{2m - 1}{3} = \frac{5}{2}
\][/tex]
Cross-multiplying:
[tex]\[
2(2m - 1) = 3 \cdot 5
\][/tex]
[tex]\[
4m - 2 = 15
\][/tex]
[tex]\[
4m = 17
\][/tex]
[tex]\[
m = \frac{17}{4} = 4.25
\][/tex]
Next, let's equate [tex]\(\frac{b_1}{b_2}\)[/tex] and [tex]\(\frac{c_1}{c_2}\)[/tex]:
[tex]\[
\frac{3}{n - 1} = \frac{5}{2}
\][/tex]
Cross-multiplying:
[tex]\[
3 \cdot 2 = 5 \cdot (n - 1)
\][/tex]
[tex]\[
6 = 5n - 5
\][/tex]
[tex]\[
5n = 11
\][/tex]
[tex]\[
n = \frac{11}{5} = 2.2
\][/tex]
Therefore, the values of [tex]\( m \)[/tex] and [tex]\( n \)[/tex] that make the given pair of linear equations have an infinite number of solutions are:
[tex]\[
m = 4.25 \quad \text{and} \quad n = 2.2
\][/tex]
