To determine the value of [tex]\( k \)[/tex] that will produce a system with an infinite number of solutions for the given linear system:
[tex]\[
\begin{array}{c}
-2x + 5y = 23 \\
-10x + ky = 115
\end{array}
\][/tex]
we need to compare the coefficients of the equations. For the system to have an infinite number of solutions, the ratios of the corresponding coefficients must be the same.
The system of equations is:
1. [tex]\(-2x + 5y = 23\)[/tex]
2. [tex]\(-10x + ky = 115\)[/tex]
For these two equations to be proportional (i.e., to represent the same line and thus have infinitely many solutions), the ratios of the coefficients must satisfy the following condition:
[tex]\[
\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}
\][/tex]
Here, the coefficients are:
- For the first equation [tex]\(-2x + 5y = 23\)[/tex]:
- [tex]\(a_1 = -2\)[/tex]
- [tex]\(b_1 = 5\)[/tex]
- [tex]\(c_1 = 23\)[/tex]
- For the second equation [tex]\(-10x + ky = 115\)[/tex]:
- [tex]\(a_2 = -10\)[/tex]
- [tex]\(b_2 = k\)[/tex]
- [tex]\(c_2 = 115\)[/tex]
Firstly, we compare the coefficients [tex]\( \frac{a_1}{a_2} \)[/tex]:
[tex]\[
\frac{-2}{-10} = \frac{1}{5}
\][/tex]
Next, we will set this equal to the ratio [tex]\( \frac{b_1}{b_2} \)[/tex]:
[tex]\[
\frac{1}{5} = \frac{5}{k}
\][/tex]
Solving for [tex]\( k \)[/tex], we multiply both sides by [tex]\( k \)[/tex]:
[tex]\[
\frac{1}{5} \cdot k = 5
\][/tex]
Then, multiply both sides by 5 to get rid of the fraction:
[tex]\[
k = 25
\][/tex]
Therefore, the value of [tex]\( k \)[/tex] that will produce a system with an infinite number of solutions is [tex]\( 25 \)[/tex].
