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Sal solves several examples where he reasons about the number of solutions of systems of equations using algebraic reasoning.
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- If both sides equal zero is that also a zero solution equation?(10 votes)
How many solutions does this system have? Pls help!(4 votes)
- This system of linear equations have only one solution.
That is because this system of equations is written in slope-intercept form:
In which m is the slope and b is the y-intercept.
So in the first equation, -2 is the slope.
And in the second equation, 3 is the slope.
And it becomes very obvious -- two lines with a DIFFERENT slope will always intersect at some point!
So this system has only one solution.
P.S. sorry for answering 9 months later!(8 votes)
- I'm purposefully trying to make a difficult question.
What is the solution to this system of linear equations?
Where m = Graham's Number
I figure it's something VERY close to (0,-2) because the two lines will almost seem to be on top of one another in an almost vertical line.
To be a little more accurate, I know the solution will have an x value slightly more negative than 0 (-0.000...1) and a y value slightly more negative than -2 (-2.000...1).
How do I show my solution algebraically?
Another question: Picking an arbitrary location on the y-axis, how would I show the distance from those two lines?(3 votes)
- I got a different answer to your first question.
Through substitution, x = 3.
So the solution to the system of equations y = mx - 1 and y = (m - 1)x - 2 is the ordered pair (3, y).
To find y, we simplify again and see that:
y = 3(Graham's Number) - 5
So the lines will intersect at (3, y) where y is an extremely big number.(1 vote)
- I have difficulty understanding this kind of equations:
even when I get a hint, the hint is talking about another equation!(5 votes)
- You have probably already figured this out by now, but you can use the second equation to solve for the first one :D Since the second equation says that y=x-3, you can substitute "x-3" into the first equation so that there's only one variable (x). Then you can solve the rest of the equation, I believe.(1 vote)
- How did you get -36 on the first problem(3 votes)
- He multiplied or divided the whole second equation by -1 which in effect just changes the signs in the whole equation such that 5x - 9y = 36 becomes - 5x +9y - - 36(4 votes)
- On the last question, Sal wrote the co-ordinates 4/5, -1 but, how does he know that it' exactly one solution?(3 votes)
- If you mean the one before the last exercise, try to think this way:
There were 2 equations:
Those 2 equations forms a system of equations right?
To validate if the system has indeed only one solution, all of the lines within the system must have a different y-intercept.
Since there is 2 equations in the system, we can say that there are 2 lines as well.
To check their y-intercept you can assume x is zero for all of them.
5(0)-2y=6 -> -2y=6 -> y=-3
5(0)+3y=1 -> 3y=1 -> y=1/3
Since their y-intercept is different, there is only 1 solution to the system.
But, there are many ways to solve something like this, what Saul did was following the exercise clue: "Albus takes several correct steps that lead to the equation 5y=-5", which means that Albus joined both lines "behaviors" (equations) and ended up with a single equation (behavior).
And if you observe carefully, you'll notice that what he did was a subtraction.
And that leads us to a different way of knowing if a system equation has one or more solutions, by solving them instead of analyzing its behaviors.
If you join the behaviors (by adding, subtracting or isolating a variable and then merging the equations) and get a result different from 0=0 it means that the equation has only one possible solution (which will be either an 'y' or a 'x', that means where the lines encounter each other, in the case of the exercise Albus gave you the 'y').(3 votes)
- At2:07, even if the 0 had an x, like 0x = -20, it still wouldn't work, right? Because everything times 0 = 0, not -20.(3 votes)
- Yes, in fact, x is undefined in that equation because to solve for it, you would have to divide by 0 which is undefined.(3 votes)
- 5x - 9y = 16
5x - 9y = 36
Both 16 and 36 are equal to 5x - 9y. This means 16 = 36, but they aren't equal. I don't understand.(2 votes)
- Which also means that the two lines created by these equations are parallel (they both have a slope of 5/9, but different y intercepts - -16/9 and -4).(3 votes)
- Sooo, this is my logic:
after solving for y, I get an equation, say K.x +C(where K is coefficeint of x and C is a constant). Then, the equations have-
1. No solutions if K is same but C is different in both equation.
2. infinite solutions if K and C are equal for both equations.
3. and, one solution if K is different.
Is this right?(2 votes)
- Yes, that is correct.
K in this instance is called the 'slope' of a line. It's the number of steps that 'y' will grow at every step of 'x'. C is the y-intercept.
When both equations are equal, you'll get infinite intersections, since the two lines overlap.
When both equations have the same slope, but not the same y-intercept, they'll be parallel to each other and no intersections means no solutions.
When both equations have different slopes than regardless of the y-intercept they'll intersect for certain, therefore it has exactly one solution.(4 votes)
- How would you know if a system has exactly one solution, and i didn't get how in5:37(4/5, 1) represents exactly one solution(3 votes)
- I'm not sure what you mean. That is the answer. You only have one place on the graph where the two lines intersect. Therefore, you only have one answer that satisfies both equations.(1 vote)
how many solutions does the following system of linear equations have and I have my system right over here there's a couple of ways to think about it one way is to think about them graphically and think about well are they the same line in which case they would have an infinite number of solutions are they parallel in which case they never intersect you'd have no solutions or do they intersect exactly one place in case you would have one solution but instead we're going to do this algebraically so let's try to actually just solve the system and see what we get so the first equation I'm going to leave that unchanged 5x minus 9y is equal to is equal to is 16 now this second equation right over here let's say I want to cancel out the X terms so let me multiply the second equation by negative one so I have a negative 5x that I can cancel out with the 5x so if I multiply the second equation by negative one I'm going to have negative 5x plus 9y plus 9 Y is equal to negative 36 negative 36 now what I'm going to do is I'm going to add the left side of the equations and the right sides of the equations to get a new equation so 5x minus 5x well that's going to be 0 negative 9y plus 9y well that's going to be 0 again I don't even have to write it it's just going to be 0 on the left-hand side and on the right-hand side I'm going to have 16 minus 36 so negative 20 so now I'm left with the somewhat bizarre-looking equation that says that 0 is equal to negative 20 now one way you might say well how does this make any sense and the way to think about it is well are there any X Y values 4 which is 0 is going to be equal to negative 20 well no 0 is never going to be equal to negative 20 and so it doesn't matter what X Y values you can never find an X well X or XY pair that's going to make 0 equal to negative 20 in fact the X's and Y's have disappeared from this equation there's no way that this is going to be true so we have no we have no solutions now if you were to plot these if you were to plot each of these lines you would see that they are parallel lines and that's why they have the same slope different y-intercepts and that's why we have no solutions they don't intersect let's do another one of these this is this is fun all right how many solutions does the following system of linear equations have so let's do the same thing I'm going to keep the first equation the same negative 6x plus 4y is equal to 2 and the second equation let me just see if I can cancel out the X term so if I have a negative 6x if I multiply this by 2 I'm going to have a positive 6x so I can let's see I'm going to multiply this whole equation both sides of it by 2 so I'm going to have 6x 3 x times 2 is 6x negative 2y times 2 is negative 4y and that is going to be equal to negative 2 now let's do the same thing let's add the left sides and let's add the right sides so negative 6x plus 6x well that's going to be 0 4y minus 4y that's 0 we just have a 0 on the left-hand side now on the right-hand side we have 2 plus negative 2 well that's 0 so this is a little bit different it still looks a little bit bizarre 0 equals 0 last time we had 0 is equal to we add what 0 is equal to negative 20 now we have 0 equals 0 so one way to think about it is even though the X's and Y's are no longer in this equation okay well what XY pairs is it going to be true for is that it's going to make it true that 0 is equal to 0 well this is going to be true no matter what x and y are in fact x and y are not involved in this equation anymore 0 is always going to be equal to 0 so this is going to have an infinitely this is going to have infinitely many solutions here and that's because these are the same lines they just look a little bit different algebraically but if you scale one of them in the right way in fact if you just multiply both sides of this one the second one by negative 2 you're going to get the top one and so they actually represent the exact same lines you have an infinitely many solutions all right when trying to find the solution to the following system of linear equations Yvonne takes several correct steps that lead to the equation negative 5 is equal to 20 how many solutions is the system of linear equations have I don't even have to look at the system right over here the fact that she got the statement that can never be true negative 5 is never going to be equal to 20 tells us that she has no those solutions and once again if you were to plot these graphically you would see that these are parallel lines that's why they have no solutions they never intersect there's no XY pair that satisfies both of these constraints let's do a let's do a couple more of these when trying to find the solution to the following system of linear equations Alba's take several correct steps that lead to the equation 5y is equal to negative 5 and say how many solutions does this system of linear equations have well 5y equals negative 5 we could divide both sides by 5 and we get Y is equal to negative 1 and then if you substitute back in Y is equal to negative 1 if you did it in this first equation if Y is equal to negative 1 all of this becomes positive 2 you can subtract 2 from both sides and you get 5x is equal to 4 or you'd get what X is equal to 4/5 or if you put negative 1 over here you would get 5x minus 3 is equal to 1 you could add 3 to both sides and you get 5x equals 4 again X is equal to 4/5 so you have exactly one solution you would have X is equal to 4/5 y is equal to negative 1 let's do one more when trying to find the solution of the following system of linear equations Levon take several correct steps that lead to the equation 0 equals 0 so once again I don't even need to look at this over here zero equals zero is always going to be true so this is going to have an infinitely this is going to be an infinitely many solutions