+0

+5
62
2

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
Sort:

#1
+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
+18612
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

### 26 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details