Multiple Choice - Which of these can NOT be solved?

[hakim]

Hi

Which of these equations can NOT be solved? By my maths, non of them can be...? is it a trick question?

a) 4x = 2x,
which cannot be solved for obvious reasons.

b) x + 9 = x + 8,
1 = 0
this cannot be solved as the x's negate each other.

c) 3x - 5 = 2x - 5,
3x = 2x
like a) this also cannot be solved.

d) 2x- 4 = x + 4
x = 0 (?)
this the only one that feel like has some sort of resolution..

e) 2x - 10 = 2x - 11
2x - 2x = -1
this also feels like a no.

f) 2x - 3x = 0
unless x=0 then again, cannot be solved.

SO, by my (simple) maths, non of them cannot be solved. Open for discussion. This is a question from a Swedish Exam Paper btw.

 

[wjm11]

Hi

Which of these equations can NOT be solved? By my maths, non of them can be...? is it a trick question?

a) 4x = 2x,
which cannot be solved for obvious reasons.

b) x + 9 = x + 8,
1 = 0
this cannot be solved as the x's negate each other.

c) 3x - 5 = 2x - 5,
3x = 2x
like a) this also cannot be solved.

d) 2x- 4 = x + 4
x = 0 (?)
this the only one that feel like has some sort of resolution..

e) 2x - 10 = 2x - 11
2x - 2x = -1
this also feels like a no.

f) 2x - 3x = 0
unless x=0 then again, cannot be solved.

SO, by my (simple) maths, non of them cannot be solved. Open for discussion. This is a question from a Swedish Exam Paper btw.

Why are you discarding x=0 as a solution? That IS the solution to several of these problems. It's the ones that give a nonsense answer, such as 1 = 0 that have no solution.

 

[hakim]

Why are you discarding x=0 as a solution? That IS the solution to several of these problems. It's the ones that give a nonsense answer, such as 1 = 0 that have no solution.

So then only d) and f) can be solved?

 

[wjm11]

So then only d) and f) can be solved?

So, what about a) and c)? Did you try x = 0 in them?

 

[hakim]

So, what about a) and c)? Did you try x = 0 in them?

OK, but with regards to algebra - how do you solve: 4x = 2x OR 3x = 2x ?

 

[wjm11]

OK, but with regards to algebra - how do you solve: 4x = 2x OR 3x = 2x ?

4x = 2x
Subtract 2x from each side:
4x - 2x = 2x - 2x
2x = 0
Divide both sides by 2:
x = 0.

Similar approach for the other problem.

 

[stapel]

a) 4x = 2x

So you have:

. . . . .4x = 2x

. . . . .4x - 2x = 2x - 2x

. . . . .2x = 0

...and so forth.

which cannot be solved for obvious reasons.

What would be those "obvious reasons"?

b) x + 9 = x + 8,
1 = 0
this cannot be solved as the x's negate each other.

No. This cannot be solved because there is no possible x-value that could possibly make one equal to zero. It is not because the variable disappears. For instance:

. . . . .Solve:
. . . . .x + 5 = 2(4 - x) - 3(1 - x)

. . . . .Solution:
. . . . .x + 5 = 2(4) + 2(-x) - 3(1) - 3(-x)

. . . . .x + 5 = 8 - 2x - 3 + 3x

. . . . .x + 5 = 8 - 3 - 2x + 3x

. . . . .x + 5 = 5 + x

. . . . .x - x + 5 = 5 + x - x

. . . . .5 = 5

The variable disappears, but five is still (and always) equal to five. So any value of x would work. In this sort of equation, the solution is "all x" or "all real numbers".

c) 3x - 5 = 2x - 5,
3x = 2x
like a) this also cannot be solved.

Wrong. Like (a), this can be solved.

d) 2x- 4 = x + 4

. . . . .2x - x - 4 = x - x + 4

. . . . .x - 4 = 0 + 4

. . . . .x - 4 + 4 = 4 + 4

...and so forth.

x = 0 (?)

What were your steps? How did you arrive that this answer?

e) 2x - 10 = 2x - 11
2x - 2x = -1
this also feels like a no.

What do you mean by "feels like a no". What do you get when you simplify the left-hand side? What does this mean, logically?

f) 2x - 3x = 0
unless x=0 then again, cannot be solved.

Do you think that zero is not actually a number, so therefore it cannot be a solution value? :shock:

 

[Steven G]

Hi

Which of these equations can NOT be solved? By my maths, non of them can be...? is it a trick question?

a) 4x = 2x,
which cannot be solved for obvious reasons.

b) x + 9 = x + 8,
1 = 0
this cannot be solved as the x's negate each other.

c) 3x - 5 = 2x - 5,
3x = 2x
like a) this also cannot be solved.

d) 2x- 4 = x + 4
x = 0 (?)
this the only one that feel like has some sort of resolution..

e) 2x - 10 = 2x - 11
2x - 2x = -1
this also feels like a no.

f) 2x - 3x = 0
unless x=0 then again, cannot be solved.

SO, by my (simple) maths, non of them cannot be solved. Open for discussion. This is a question from a Swedish Exam Paper btw.

Different multiples of 0 are 0.
7*0 = 4*0 = -3*0 = .... =0.