# Write a system of equations in 11 unknowns tomb

A very simple example of that is, well, one, they could all be parallel to each other.

## Cramers rule determinant

And here we have three equations with three unknowns. How much did John invest in each type of fund? How much did he invest in each type of fund? A very simple example of that is, well, one, they could all be parallel to each other. This should be equal to negative 10, and it is. And if we want to eliminate the y's, we can just add these two equations. And then we could use these two equations to come up with another equation that'll only be in terms of x and z. And just so you have a way to visualize this, each of these equations would actually be a plane in three dimensions. An infinite number of solutions can result from several situations. Let me write it over here. Negative y plus y cancels out.

Then another plane maybe pops out like that, goes underneath. So you see, this kind of forms a triangle, and they don't all intersect in one point.

Analysis of the Solution In this system, each plane intersects the other two, but not at the same location. Graphically, a system with no solution is represented by three planes with no point in common.

Maybe it intersects over here and over here, and so it pops out like that. When a system is dependent, we can find general expressions for the solutions. And so the intersection of this plane, the x, y, and z-coordinates that would satisfy all three of these constraints the way I drew them, would be right over here.

So we have 2 times x, 2 times negative 2 minus z is equal to negative 7, or negative 4 minus z is equal to negative 7.

Negative 4 plus y, so plus 1 minus z, so minus 3 needs to be equal to negative 6. So if I add these two equations, I get 3x plus z is equal to negative 3.

### Simultaneous equations questions

And then negative 3z plus 2z, that gives us just a negative z. How To: Given a linear system of three equations, solve for three unknowns. So maybe one plane looks like that. And then it goes below it like that, and it goes like that. And so the intersection of this plane, the x, y, and z-coordinates that would satisfy all three of these constraints the way I drew them, would be right over here. A very simple example of that is, well, one, they could all be parallel to each other. Graphically, a system with no solution is represented by three planes with no point in common. Well, we know z is minus 3 times 3 should be equal to negative Negative 4 plus 1 is negative 3, and then you subtract 3 again. I'm just doing this for visualization purposes. So we have 2 times x, 2 times negative 2 minus z is equal to negative 7, or negative 4 minus z is equal to negative 7. Write the result as row 2. An infinite number of solutions can result from several situations. This is a little bit more traditional of a problem.

And maybe this plane over here, it intersects right over there, and it comes popping out like this. So that's what we're looking for.

So we have x minus y plus 2z is equal to 3, and we have 2x plus y minus z is equal to negative 6. A system of equations in three variables is dependent if it has an infinite number of solutions.

## Cramers rule calculator

Pick any pair of equations and solve for one variable. Show Solution To solve this problem, we use all of the information given and set up three equations. Provided by: Lumen Learning. And here we have three equations with three unknowns. But you can imagine if I were to draw three-dimensional space over here. Systems of equations in three variables that are dependent could result from three identical planes, three planes intersecting at a line, or two identical planes that intersect the third on a line. And then negative 3z plus 2z, that gives us just a negative z.

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