High School Attempt to solve this system of three linear equations

[Elara04]

TL;DR

Is it possible to add an unknown or co-efficient to make an unsolvable system of equation solvable???

The point (1, 5) is on the curve: y=ax^2+bx+c. This point gives the linear equation: 5 = a + b + c. A second point on the curve, (2, 10) gives the linear equation 10=4a+2b+c. A student called Erika thinks that the point (2, 19) is also on the curve.

5 = a + b + c.
10=4a+2b+c
19=4a+2b+c

the system of equations is unsolvable, but I have been told that there is a way to add another unknown or co-efficient to make it possible to answer the question. Does anyone have any ideas?

 

[berkeman]

Summary: Is it possible to add an unknown or co-efficient to make an unsolvable system of equation solvable?

The point (1, 5) is on the curve: y=ax^2+bx+c. This point gives the linear equation: 5 = a + b + c. A second point on the curve, (2, 10) gives the linear equation 10=4a+2b+c. A student called Erika thinks that the point (2, 19) is also on the curve.

5 = a + b + c.
10=4a+2b+c
19=4a+2b+c

the system of equations is unsolvable, but I have been told that there is a way to add another unknown or co-efficient to make it possible to answer the question. Does anyone have any ideas?

Welcome to PF.

Why do you say this system is unsolvable? You have 3 equations and 3 unknowns -- are the equations not independent?

Please tell us more about what you know about solving simulataneous equations so we can try to help your understanding.

Also, is this for schoolwork?

 

[berkeman]

Why do you say this system is unsolvable? You have 3 equations and 3 unknowns -- are the equations not independent?

10=4a+2b+c
19=4a+2b+c

Oh, LOL. Are you trolling us?

 

[topsquark]

Summary: Is it possible to add an unknown or co-efficient to make an unsolvable system of equation solvable?

The point (1, 5) is on the curve: y=ax^2+bx+c. This point gives the linear equation: 5 = a + b + c. A second point on the curve, (2, 10) gives the linear equation 10=4a+2b+c. A student called Erika thinks that the point (2, 19) is also on the curve.

5 = a + b + c.
10=4a+2b+c
19=4a+2b+c

the system of equations is unsolvable, but I have been told that there is a way to add another unknown or co-efficient to make it possible to answer the question. Does anyone have any ideas?

Don't think about the equations, think about the function $y = ax^2 + bx + c$. Can we have both (2, 10) and (2, 19) on this curve?

-Dan

 

[Mark44]

Why do you say this system is unsolvable? You have 3 equations and 3 unknowns

Which isn't enough to make the system solvable.
10=4a+2b+c
19=4a+2b+c
The 2nd and 3rd equations are inconsistent. No set of values a, b, and c can satisfy both equations. Geometrically, these two equations represent two parallel planes in $\mathbb R^3$. Obviously, two parallel planes can't share any common points.

Oh, LOL. Are you trolling us?

No, not at all. The 3rd equation comes from a hypothetical student Erika who thinks that a certain point lies on the curve.

 

[DaveE]

Typo maybe? It just doesn't make sense.

Anyway a simple linear set of equations that is "unsolvable" is unsolvable. The point is "unsolvable" in this context doesn't mean "I don't know the answer" it means "there isn't an answer, and there never will be".

You certainly could change the question to make a different problem solvable though.

 

[DaveE]

The 3rd equation comes from a hypothetical student Erika who thinks that a certain point lies on the curve.

Hypothetical Erika is a waste of time. State the problem clearly and proceed with the analysis, the back story isn't necessary.

BTW, this isn't calculus.

 

[Mark44]

Hypothetical Erika is a waste of time. State the problem clearly and proceed with the analysis, the back story isn't necessary.

The problem was stated clearly. The part about Erika was a necessary part of the problem.

 

[mpresic3]

Could try to find solutions modulo (?) . I would not want to try it though. eg 7 mod 5 = 2 so the same "2" could represent two different numbers.

 

[FactChecker]

The last two equations are contradictory as they stand:
10=4a+2b+c
19=4a+2b+c

Adding another variable makes a solvable problem:
10=4a+2b+c
19=4a+2b+c + d
Clearly, we can set d = 9 and these two equations are now redundant, not contradictory, so the new system of equations is solvable.
This is not as unusual and useless as you might think. In linear programming, there are algorithms (Simplex Method) where variables (slack, surplus, and artificial variables) are added to get an easy initial solution. The algorithm then finds other solutions that satisfy the original problem which had inequalities, not equalities as constraints.

 

[pasmith]

TL;DR Summary: Is it possible to add an unknown or co-efficient to make an unsolvable system of equation solvable???

The point (1, 5) is on the curve: y=ax^2+bx+c. This point gives the linear equation: 5 = a + b + c. A second point on the curve, (2, 10) gives the linear equation 10=4a+2b+c. A student called Erika thinks that the point (2, 19) is also on the curve.

5 = a + b + c.
10=4a+2b+c
19=4a+2b+c

the system of equations is unsolvable, but I have been told that there is a way to add another unknown or co-efficient to make it possible to answer the question. Does anyone have any ideas?

If the curve has an equation of the form y = f(x) then for each x there exists at most one y such that (x,y) is on the curve. It is given that (2,10) is on the curve, so (2,19) is not on the curve and Erika is wrong. Alternatively, it may be that the assumption that the curve has an equation of the form y = f(x) is incorrect, and Erika is right: as for example if x = g(y) = \frac{-y^2 + 29y - 50}{70}.