Help w/ systems of Linear Equations?

Question

Problem 1

Determine the value of k for which the system of linear equations
4x + 5y = 12
8x + ky = 13 has no solution.

AND

Problem 2

Determine the value of k for which the system of linear equations
4x + 5y = 12
x + ky = 3 has infinitely many solutions.

I need help. I'm stumped.

Answer 1

4x + 5y = 12
8x + ky = 13

4x + 5y = 12
5y = -4x + 12
y = (-4/5)x + (12/5)

8x + ky = 13
ky = -8x + 13
y = (-8/k)x + (13/k)

for there to be no solution, the equations must have the exact same slope and different y-intercepts.

(-8/k) = (-4/5)
-4k = -40
k = 10

ANS : k = 10

------------------------------------

4x + 5y = 12
5y = -4x + 12
y = (-4/5)x + (12/5)

x + ky = 3
ky = -x + 3
y = (-x/k) + (3/k)

same slope and same y-intercept

(-4/5)x = (-x/k)
-4xk = -5x
-4k = -5
k = (5/4)

if you put it (5/4) for k

x + (5/4)y = 3

this would become

4x + 5y = 12

As you can see, they are both are identical and therefore have infinate solutions.

Answer 2

For the first one, if the two equations represent lines on a graph, and we make those lines parallel and distinct, there wlil be no place where the lines meet, and hence no solution.

The slope of

4x + 5y = 12 is -4/5

We can make the slope of

8x + ky = 13 the same by letting k = 10. That is your answer.

For the second pair, we want to two lines to be indentical, with the same slope and intercept. Obviously,

x + (5/4)y = 3

is identical to

4x + 5y = 12

Your answer is k = 5/4

Answer 3

A general trick is to make the determinant of the coefficient matrix equal to zero.

In the case of two equations in two unknowns, you "cross multiply" the two factors and make there difference zero.

Problem 1: 4 * k - 5 * 8 = 0, therefore k = 10.

Problem 2: 4 * k - 5 * 1 = 0, therefore k = 5/4.

Answer 4

For two linear equations a1x+b1y=c1 and a2x+b2y=c2

the condition for no solution is
a1/a2 = b1/b2 not=c1/c2

Putting the values of a1, b1, a2 and b2 we get

4/8 = 5/k

k=10

Prob. 2
The condition for infinite solutions is

a1/a2 = b1/b2 = c1/c2

Putting the values

4= 5/k

k=5/4