# Thread: Systems of linear equations - no solution, unique solution, infinite solutions

1. ## Systems of linear equations - no solution, unique solution, infinite solutions

I'm just gona copy these from my book:

1. x-y = 6 and 2x-2y = 12 have infinitely solutions, show this using parameters?

Obviously both equations are the same line, but the answer is x = t+6 where t = real number

I assume a 3rd variable is involved...

2. Find the values of b and c for which equations x+5y = 4 and 2x + by = c that have

a. a unique solution b. infinite set of solutions c. no solution

3. Solve simultaneous equations 2x-3y = 4 and x + ky = 2 where k is a constant, also, find value of k with no unique solution

How do I solve these

2. Originally Posted by NewtoMath
I'm just gona copy these from my book:

1. x-y = 6 and 2x-2y = 12 have infinitely solutions, show this using parameters?

Obviously both equations are the same line, but the answer is x = t+6 where t = real number

I assume a 3rd variable is involved...
The set of solutions consists of ordered pairs (x, y). If you want to calculate some elements of this set you need a parameter such that

$(x, y) = (t+6, t)$ or $(x, y) = (t, t - 6)$

2. Find the values of b and c for which equations x+5y = 4 and 2x + by = c that have

a. a unique solution b. infinite set of solutions c. no solution
Solve this system of equations considering b, c as constants. You'll get:

$x=\dfrac{5(c-8)}{b-10}+4~\wedge~y=\dfrac{c-8}{b-10}$

You now can determine
a.) $b\neq10~\wedge~c\neq8$
b.) $b=10~\wedge~c=8$
c.) $b=10~\wedge~c\neq8$

3. Solve simultaneous equations 2x-3y = 4 and x + ky = 2 where k is a constant, also, find value of k with no unique solution

How do I solve these
If you solve this system of equation at first for y you'll get y = 0 and x = 2.

So it seems as if k can have any value. But if you solve the system at first for x you'll get $x = \dfrac{4k+6}{2k+3}$

Obviously this fraction is not defined for $k = -\frac32$

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# find value of k gor which eq do not have unique solution 2x ky=1

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