+0  
 
+5
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            3

a y = - --- x

            4

4x - 5y = 0

 

For what values of does the system of equations of linear equations in the variables x and y have infinitely many solutions?

Guest Feb 27, 2017
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 #1
avatar+61 
0

start by placing both in slope-intercept form (y=mx+b)

 

so you have a(y)=-3/4x

and y=4/5x

 

in order for a system of linear equations to have infinitely many solutions the have to have the same slope and intercept, right now

-3/4 \(\neq\)4/5

 

so we have to figure what "a" needs to equal in order form the lines to be the same.

 

so  (-3/4)/x=4/5

solve for x

x=(4/5)*(-3/4)

x=-3/5

 

so we know that "A' is going to be the reciprical of x**

 

A=-5/3

 

 

**how we decided X and why we know that x= the recipricle

so ay=-3/4x

so it would end up being y= ((-3/4)/a)x

and dividing by a fraction is the same as multiplying by its reciprical.

brkr19  Feb 27, 2017
 #2
avatar+18777 
0

For what values of a does the system of equations of linear equations
in the variables x and y have infinitely many solutions?

\(\begin{array}{|rcll|} \hline ay &=& -\frac{3}{4}x \\ 4x-5y &=& 0 \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline ay &=& -\frac{3}{4}x \quad &| \quad + \frac{3}{4}x \\ \frac{3}{4}x + ay &=& 0 \quad &| \quad \cdot \frac{16}{3} \\ 4x + \frac{16}{3}ay &=& 0 \\ \hline \end{array}\)

 

infinitely many solutions, if  \(\frac{16}{3}a = -5\)

\(\begin{array}{|rcll|} \hline \frac{16}{3}a &=& -5 \quad &| \quad \cdot \frac{3}{16} \\ a &=& -5\cdot \frac{3}{16} \\ a &=& -\frac{15}{16} \\ \hline \end{array}\)

 

laugh

heureka  Feb 28, 2017

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