Comparing

Algebra Level 2

True or False?

If real numbers ${\color{#D61F06}a},\ {\color{#3D99F6}b},\ x,$ and $y$ satisfy ${\color{#D61F06}a}x + {\color{#3D99F6}5}y = {\color{#D61F06}2}x + {\color{#3D99F6}b}y,$ then it must be the case that ${\color{#D61F06}a=2} \text{ and } {\color{#3D99F6}b=5}.$

True False

7 solutions

Zach Abueg
Apr 29, 2017

They can be true, but not necessarily must be true. Counterexample:

$x = y = 0$

The problems states the IF <equation> THEN <values of a & b>.. Mr. Abueg showed that at least one solution to the equation existed other than the one stated. IF m THEN n, and something other than n was shown to be true. This falsifies the IF-THEN statement. A single counterexample is enough.

Ron Amundson - 4 years, 1 month ago

Thank you.

Zach Abueg - 4 years, 1 month ago

This question is about semantics. A little boring, actually.

Marcio Castro - 4 years, 1 month ago

the problem states that the equation is true for all real numbers, which means for all x and y. The counterexample only works for specific x and y.

Walter Neilsen-Steinhardt - 4 years, 1 month ago

If it can be True, should be true at least for this specific set of values

Gopal Prashanth Muthyala - 4 years, 1 month ago

I got this wrong, although I am still struggling with why. I think instead of focusing on the mathematics, the focus should have been on pure logic. The mathematics appears to be right only in this instance? The if..then part of the equation adds conditions that should lead to a logical sequence. If this is true, then that is true. The real answer is maybe. Why? Because there are other possibilities, since the logic is if...then.

Brenda Gaines Hunter - 4 years, 1 month ago

"...which means that the statement could be true for any values of a or b. "

Richard Desper - 4 years, 1 month ago

That is a trick question. It can be true, yes, and that is the thing I based my answer on.

gh dgfhdgf - 3 years, 3 months ago

To Walter - although the counterexample works for the specific x and y mentioned, "all x and y" means that the statement must hold true for every possible value of x and y. Finding a single counterexample is enough to invalidate the statement, as it's now only true for "all x and y, except..."

EB Z - 4 years, 1 month ago

Amartya Harsh Singh
May 8, 2017

$ax+5y=2x+by$ then $(a-2)x+(5-b)y=0$ , and thus we get another equation $a'x+b'y=0$ , which is another straight line with an infinite number of points on it; and hence infinite solutions.

Moderator note:

To some extent, this problem attempts to trick the solver into thinking free variables are bound variables.

Consider: If $x + 5 = 2 + 5 ,$ then $x = 2$ is a true statement; $x$ only has one possible value, and is a bound variable.

However, if $x + 5 = 2 + y,$ it is not true that $x = 5$ and $y = 3;$ both variables are now free and the graph of the equation forms a line with a positive slope where each point on the line is one of an infinite number of solutions.

Adding the extra variables $a$ and $b$ adds even more freedom; $ax + 5y = 2x + by$ is a four-dimensional version of a hyperbolic paraboloid (two slices of the four-dimensional shape are shown below).

I'm not very sure about the straight line. In this problem, do we treat $a$ and $b$ as constants?

Christopher Boo - 4 years, 1 month ago

You've got one equation with four variables. That's not a line, it's a three-dimensional space.

Richard Desper - 4 years, 1 month ago

well it is a line you either take one ordered pair of (a,b) or (x,y) u get linear eq in 2 variables which is a lin

Rachit Pulhani - 4 years, 1 month ago

Again, that's not justify the confusion because of the sentence "... must be the case...". Maybe the options to answer should be "ternary" ("true", "false", "must be") instead of just "true" or "false"...

Marcio Castro - 4 years ago

Gary Scarr
May 8, 2017

1 equation with 4 unknowns so can't have a single solution, so must be False

while it is not necessarily true, that is not the way the question was worded. True or false without a qualifier allows those numbers as a possible correct solution and the answer would be true.

Joseph Nicastro - 4 years, 1 month ago

If x & y are variables than this must be true. However, since we only require them to be real numbers than this is not necessarily true.

Douglas Johnson - 4 years, 1 month ago

@Joseph Nicastro, the question states 'must be true'

Chris Wood - 4 years, 1 month ago

'Must be the case' rather..

Chris Wood - 4 years, 1 month ago

Saurabh Chaturvedi
May 9, 2017

This would simply give (a-2)x + (5-b)y = 0. For different values of x and y, we would get different values of a and b. What if we draw a vector field where at every point (x, y) we draw a vector (a, b) such that the above equation holds?

I don't understand your solution. Why are we discussing vectors in this context?

Christopher Boo - 4 years, 1 month ago

I Jones
May 9, 2017

If any other solution is true the statement is automatically false.
For example if x and y are both 0 then a and b can be any number.

Right, the equation can have solutions for $x$ and $y$ even when $a\neq 2$ and $b \neq 5$ .

Pranshu Gaba - 4 years, 1 month ago

Hetansh Mehta
May 8, 2017

above equation renders, x/y=(b-5)/(a-2), thus a=2 and b=5 yields 0/0 which is undefined

It does not render that, since you either use the general value a for a, in which case it is undefined, except when it is a=2, or you use the a=2 equation to begin with, and then you can't do the /(a-2), because you know it is an illegal move. Case in point: you can't just add additional information in the middle of a problem, you either know it from the start, or not.

József Inczefi - 4 years, 1 month ago

What the author of the problem suggest by saying that the trivial solution (a=2 b=5)is WRONG by making others to come to include undetermination is exactly the opposite of what you do when you try to eliminate an undetermination. So do not be too smart!

Dinu Fotescu - 4 years, 1 month ago

Gabriel Souza
May 12, 2017

Simply assume that x = y = 1.

Couldn't a = 1 and b = 4? That's a different answer already, so the statement is false, for there are multiple right answers

Comparing

7 solutions

Moderator note:

0 pending reports