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{:[ay=(3)/(4)x],[x-5y=0]:}
For what value of
a does the system of linear equations in the variables
x and
y have infinitely many solutions?

$\begin{aligned} a y & =\frac{3}{4} x \\ x-5 y & =0 \end{aligned}$ $\newline$ For what value of $a$ does the system of linear equations in the variables $x$ and $y$ have infinitely many solutions?

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Domain and range of absolute value functions: equations

Full solution

Q.

\begin{aligned} a y & =\frac{3}{4} x \\ x-5 y & =0 \end{aligned}

\newline

For what value of

a

does the system of linear equations in the variables

x

and

y

have infinitely many solutions?

Given System of Equations: We are given a system of two linear equations: $\newline$ $1$ ) $ay = \left(\frac{3}{4}\right)x$ $\newline$ $2$ ) $x - 5y = 0$ $\newline$ To have infinitely many solutions, both equations must represent the same line. This means that the coefficients of $x$ and $y$ in both equations must be proportional.
Expressing Equation $2$ in Slope-Intercept Form: Let's express equation $2$ ) in the form of $y = mx$ , where $m$ is the slope of the line: $\newline$ $x - 5y = 0$ $\newline$ $5y = x$ $\newline$ $y = \left(\frac{1}{5}\right)x$ $\newline$ Now we have the slope of the second line, which is $\frac{1}{5}$ .
Setting the Slopes Equal: For the system to have infinitely many solutions, the slope of the first equation must be equal to the slope of the second equation. Therefore, we set the slopes equal to each other: $\newline$ $\frac{3}{4}$ = $a \cdot \frac{1}{5}$
Solving for a: Now we solve for a: $\newline$ $a = \frac{3}{4} \times \frac{5}{1}$ $\newline$ $a = \frac{15}{4}$ $\newline$ $a = 3.75$

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What is the domain of this function?

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