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{:[8(x-9y)=8-6(x-4)],[4x-Ay=16-3x]:}
In the system of equations,
A is a constant. For what value of
A does the system of linear equations have infinitely many solutions?

$\begin{array}{c} 8(x-9 y)=8-6(x-4) \\ 4 x-A y=16-3 x \end{array}$ $\newline$ In the system of equations, $A$ is a constant. For what value of $A$ does the system of linear equations have infinitely many solutions?

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Solve rational equations

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Q.

\begin{array}{c} 8(x-9 y)=8-6(x-4) \\ 4 x-A y=16-3 x \end{array}

\newline

In the system of equations,

A

is a constant. For what value of

A

does the system of linear equations have infinitely many solutions?

Simplify First Equation: First, let's simplify the first equation in the system. $\newline$ $8(x - 9y) = 8 - 6(x - 4)$ $\newline$ Distribute the $8$ and $-6$ on both sides of the equation. $\newline$ $8x - 72y = 8 - 6x + 24$ $\newline$ Combine like terms. $\newline$ $8x - 72y + 6x = 8 + 24$ $\newline$ $14x - 72y = 32$ $\newline$ Now, divide the entire equation by $2$ to simplify it further. $\newline$ $(14x - 72y) / 2 = 32 / 2$ $\newline$ $7x - 36y = 16$
Simplify Second Equation: Next, let's simplify the second equation in the system. $\newline$ $4x - Ay = 16 - 3x$ $\newline$ Add $3x$ to both sides of the equation to move all x terms to one side. $\newline$ $4x + 3x - Ay = 16$ $\newline$ $7x - Ay = 16$
Check for Infinitely Many Solutions: For the system to have infinitely many solutions, the two equations must be the same, or one must be a multiple of the other. This means that the coefficients of $x$ and $y$ must be proportional, and the constants on the right side of the equation must also be the same. $\newline$ From the simplified equations, we have: $\newline$ $7x - 36y = 16$ (from the first equation) $\newline$ $7x - Ay = 16$ (from the second equation) $\newline$ For these to be the same, $A$ must be equal to $36$ .

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