In each case, how many such equations can you find? Solution While it is possible to solve this problem purely algebraically, thinking about how the question relates to the graphs of the equations makes the problem much easier to solve. There are many possible ways to do this. However, this is not possible.

Number of solutions algebra Video transcript Determine the number of solutions for each of these equations, and they give us three equations right over here. And before I deal with these equations in particular, let's just remind ourselves about when we might have one or infinite or no solutions.

You're going to have one solution if you can, by solving the equation, come up with something like x is equal to some number.

Let's say x is equal to-- if I want to say the abstract-- x is equal to a. Or if we actually were to solve it, we'd get something like x equals 5 or 10 or negative pi-- whatever it might be.

But if you could actually solve for a specific x, then you have one solution. So this is one solution, just like that. Now if you go and you try to manipulate these equations in completely legitimate ways, but you end up with something crazy like 3 equals 5, then you have no solutions.

And if you just think about it reasonably, all of these equations are about finding an x that satisfies this. And if you were to just keep simplifying it, and you were to get something like 3 equals 5, and you were to ask yourself the question is there any x that can somehow magically make 3 equal 5, no.

No x can magically make 3 equal 5, so there's no way that you could make this thing be actually true, no matter which x you pick. So if you get something very strange like this, this means there's no solution.

On the other hand, if you get something like 5 equals and I'm just over using the number 5. It didn't have to be the number 5. It could be 7 or 10 orwhatever. And actually let me just not use 5, just to make sure that you don't think it's only for 5. If I just get something, that something is equal to itself, which is just going to be true no matter what x you pick, any x you pick, this would be true for.

Well, then you have an infinite solutions. So with that as a little bit of a primer, let's try to tackle these three equations. So over here, let's see.

Maybe we could subtract. If we want to get rid of this 2 here on the left hand side, we could subtract 2 from both sides. If we subtract 2 from both sides, we are going to be left with-- on the left hand side we're going to be left with negative 7x. And on the right hand side, you're going to be left with 2x.

This is going to cancel minus 9x. You get negative 7x is equal to negative 7x. And you probably see where this is going.

Systems of Linear Equations: No solutions and infinitely many solutions |
Elementary Row Operations Multiply one row by a nonzero number. Add a multiple of one row to a different row. |

Linear Equations |
Systems of Linear Equations: No solutions and infinitely many solutions Example 1: |

Algebra - More on the Augmented Matrix |
Number of solutions algebra Video transcript We're asked to use the drop-down to form a linear equation with infinitely many solutions. |

Algebra - More on the Augmented Matrix |
GO Consistent and Inconsistent Systems of Equations All the systems of equations that we have seen in this section so far have had unique solutions. |

This is already true for any x that you pick. Negative 7 times that x is going to be equal to negative 7 times that x. So we already are going into this scenario. But you're like hey, so I don't see 13 equals Well, what if you did something like you divide both sides by negative 7.

At this point, what I'm doing is kind of unnecessary. You already understand that negative 7 times some number is always going to be negative 7 times that number. But if we were to do this, we would get x is equal to x, and then we could subtract x from both sides.

And then you would get zero equals zero, which is true for any x that you pick. Zero is always going to be equal to zero. So any of these statements are going to be true for any x you pick.

So for this equation right over here, we have an infinite number of solutions.Systems of Linear Equations: No solutions and infinitely many solutions Example 1: A system with no solutions Consider the system of equations \[ \begin{cases} \begin{array}{ccccc} 3x & - & 2y & = & 3 \\ 6x & + & k\cdot y & = & 4 \\ \end{array} \end{cases} \] Find .

This agrees with Theorem B above, which states that a linear system with fewer equations than unknowns, if consistent, has infinitely many solutions. The condition “fewer equations than unknowns” means that the number of rows in the coefficient matrix is less than the number of unknowns.

A system of equations has infinitely many solutions. If 2y – 4x = 6 is one of the equations, which could be th Get the answers you need, now!5/5(25). A System of Linear Equations is when we have two or more linear equations working together.

Jan 21, · Write a system of equations that has infinitely many solutions, and explain why? I have a take home algebra test and I need to get a % I really need help. That's all the information the question gave attheheels.com: Resolved. So this system has infinitely many solutions, as the equations both correspond to the same line and lines have infinitely many points.

(Bonus) In parts (a), (c), and (d), there are infinitely many equations that can be found.

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Consistent and Inconsistent Systems of Equations | Wyzant Resources