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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?

ayamp;=34xx5yamp;=0 \begin{aligned} a y & =\frac{3}{4} x \\ x-5 y & =0 \end{aligned} \newlineFor what value of a a does the system of linear equations in the variables x x and y y have infinitely many solutions?

Full solution

Q. ay=34xx5y=0 \begin{aligned} a y & =\frac{3}{4} x \\ x-5 y & =0 \end{aligned} \newlineFor what value of a a does the system of linear equations in the variables x x and y y have infinitely many solutions?
  1. Write Equations: First, let's write down the system of equations given:\newline11) ay=(34)xay = \left(\frac{3}{4}\right)x\newline22) x5y=0x - 5y = 0\newlineWe are asked to find the value of aa for which this system has infinitely many solutions. This happens when the two equations are essentially the same, meaning they represent the same line.
  2. Match Equations: To compare the two equations, we need to express them in the same form. Let's rewrite the second equation to match the form of the first equation:\newlinex5y=0x - 5y = 0\newlinex=5y\Rightarrow x = 5y\newlineNow, we can write this as:\newline(55)y=(15)x\left(\frac{5}{5}\right)y = \left(\frac{1}{5}\right)x\newliney=(15)x\Rightarrow y = \left(\frac{1}{5}\right)x
  3. Find Value of a: Now we have the two equations in comparable forms:\newline11) ay=34xay = \frac{3}{4}x\newline22) y=15xy = \frac{1}{5}x\newlineFor the system to have infinitely many solutions, the coefficients of yy and xx must be the same in both equations. Therefore, we need to find the value of aa such that:\newlinea=15a = \frac{1}{5}
  4. Final Solution: Since we have determined that aa must equal 15\frac{1}{5} for the two equations to represent the same line, we have found the value of aa for which the system has infinitely many solutions.

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