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{:[2y+16=10 x],[4x-Ly=8-x]:}
In the system of equations,
L is a constant. For what value of
L does the system of linear equations have infinitely many solutions?

$\begin{array}{c} 2 y+16=10 x \\ 4 x-L y=8-x \end{array}$ $\newline$ In the system of equations, $L$ is a constant. For what value of $L$ does the system of linear equations have infinitely many solutions?

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Find the number of solutions to a system of equations

Full solution

Q.

\begin{array}{c} 2 y+16=10 x \\ 4 x-L y=8-x \end{array}

\newline

In the system of equations,

L

is a constant. For what value of

L

does the system of linear equations have infinitely many solutions?

Write Equations: First, let's write down the system of equations: $\newline$ $\begin{cases} 2y + 16 = 10x \\ 4x - Ly = 8 - x \end{cases}$ $\newline$ To have infinitely many solutions, the two equations must be dependent, meaning one is a multiple of the other.
Solve for y: Let's solve the first equation for $y$ to get it in the form $y = mx + b$ , where $m$ is the slope and $b$ is the y-intercept. $\newline$ $2y = 10x - 16$ $\newline$ $y = 5x - 8$
Express Second Equation: Now, let's express the second equation in terms of $y$ as well: $\newline$ $4x - Ly = 8 - x$ $\newline$ $Ly = 4x - 8 + x$ $\newline$ $Ly = 5x - 8$
Check Coefficients: For the system to have infinitely many solutions, the equations must represent the same line. Therefore, the coefficients of $x$ and the constants must be the same in both equations. From the first equation, we have the coefficient of $x$ as $5$ and the constant as $-8$ . From the second equation, we have the coefficient of $x$ as $5$ (which matches) and the constant as $-8$ (which also matches). The coefficient of $y$ in the second equation must be $1$ for the equations to be the same, since in the first equation, the coefficient of $y$ is implicitly $1$ . $\newline$ $L = 1$

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