Question

Can someone explain how to determine whether a system has one solution, infinitely many solutions, or no solution?

1. 6y = -5x+24
5x+3y=12

2. x=-7y +34
x+7y=32

3. 1.5x +2y =11
3x+6y=22

Answer 1

the intersect, they are the same line L2 = k* L1, or they are parallel

Answer 2

what what... I'm confused

Answer 3

*wait

Answer 4

A system will have 1 solution if your system contains n unknowns and n linearly independent equations (none of the equations can be generated by adding together multiples of other equations).

For example,
x + y = 3
3x + y = 7

You solve this by adding, subtracting, or substituting to isolate each variable:

Using subtraction:
3x + y - (x + y) = 7 - 3
3x + y - x - y = 4
2x = 4
x = 2

x + y = 3
2 + y = 3
y = 1

Using substitution:
x + y = 3
y = 3 - x

3x + y = 7
3x + (3 - x) = 7
3x + 3 - x = 7
2x + 3 = 7
2x = 4
x = 2

x + y = 3
2 + y = 3
y = 1

Answer 5

if you solve the system can can get.....

ONE ---you get a unique point (x,y),

infinity ---you get a thing like x + 5 = x + 5, that works for all x values, infinite solns, same line really

ZERO --- similar to the last one, but it will be false statememnt instead, maybe x+2 = x+3
never, parallel lines

Answer 6

oh okay... i think i get it, thanks both of you.
after i complete this, do you mind checking my answers?

Answer 7

if you look at one line, and the x coefficient and y coefficient , from the first to the next line, are both changed byu the same constant multiple , then parallel lines

Answer 8

then further, if the constant on the right is also the same multiple, same lines

Answer 9

as a quick check to save wasted work

Answer 10

oh got it thanks (:

Answer 11

A system will have infinitely many solutions if one or more of the equations in your system is a linear combination of others. For example:

x + y + z = 6
2x + 2y + z = 9
4x + 4y + 3z = 21

Since 4x + 4y + 3z = 21 is the same as 2(x + y + z) + (2x + 2y + z) = 2(6) + 9, this system will have infinitely many solutions, since you don't have enough unique equations to solve for all 3 variables. If you try to solve this system, you'll get either a tautology like 2 = 2, or two or more of your variables will be defined relative to another variable, like x = 2z, y = 3z.

Let's solve this one:

2x + 2y + z - 2(x + y + z) = 9 - 2*6
2x + 2y + 2z - 2x - 2y - 2z = 9 - 12
-z = -3
z = 3

x + y + z = 6
x + y + 3 = 6
x + y = 3
x = 3 - y

4x + 4y + 3z = 21
4(3 - y) + 4y + 3*3 = 21
12 - 4y + 4y + 9 = 21
21 = 21

As you can see, for any pair of values x and y such that x = 3 - y, this system holds true,

Answer 12

perpendicular if asked, have the products of the x coefficients + product of Y coefficients = 0, from vectors, but that is anothe rshortcut

Answer 13

A system will have infinitely many solutions if one or more of the equations in your system is a linear combination of others, but with a different answer. For obvious reasons, the following has no solutions:
x + y = 2
2x + 2y = 7

Answer 14

if one is a constant multiple of the other equation

2x + y = 5
4x + 2y = 10

same lines, just multiplied the first by 2

Answer 15

or parallel, if just the x and y were multiples, not the right side

2x + y = 5
4x + 2y = 11

almost the same line, but different multiple on the right, here it is still parallel though

Answer 16

2.
x + 7y = 34
x +7y = 32

the parallel lines thing, the x and y are a multiple, here just 1, but the other is not 1*34, so shifted, parallel lines