3

*a y =* - --- x

4

4x - 5y = 0

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

Guest Feb 27, 2017

#1**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**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 ** \(\frac{16}{3}a = -5\)**

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

heureka
Feb 28, 2017