Is It True Because of Symmetry?

Algebra Level 2

$\large {\color{#69047E}a}{\color{#D61F06}x} + {\color{#20A900}5}{\color{#69047E}b} = {\color{#D61F06}2}{\color{#69047E}a} + {\color{#69047E}b}{\color{#20A900}q} \\\\\\ \large \text{implies} \\\\\\ \large {\color{#D61F06}x} = {\color{#D61F06}2} \text{ and } {\color{#20A900}q}={\color{#20A900}5}.$

True or False?

For all real numbers ${\color{#69047E}a}, {\color{#69047E}b}, {\color{#D61F06}x}, \text{ and } {\color{#20A900}q}$ , the statement above must be true.

True False

8 solutions

James Godfhrey Saguid
Dec 19, 2015

Since it says "for all real numbers", if you let a = 0, you will see that x can be arbitrarily chosen

No no matter 2 is a real number and true for this condition so this statement is true

Mohamed Hasnaoui - 2 years, 4 months ago

Oleg Yovanovich

I just can not agree with your answer. Whatever the real values I had taken for a,b, x, and q, the statement, i.e. ax + 5b = 2a + bq, was always satisfied in its equality.

Oleg Yovanovich - 1 year, 7 months ago

I have to agree with you Oleg and Mohamed Husnaoui, possibly others, the issue I think is the meaning of "implies"/"=>". I haven't had a FOL course, but I think you have to interpret A=>B as: "From a set of facts proposed by A we claim that a subset of these facts, namely B, must also be true". This is central for me because the leading answers seems to be that "Since there are other answers, the implication doesn't hold.". If my understanding of implications is correct (which I can't prove because I can't find the corresponding literature), then this argument doesn't hold anymore, and the statement is true.

I'd love it if someone could point me to a correct definition of the implication arrow.

Ilija Kocić - 1 year, 4 months ago

Consider: we can rewrite the equation as ax - 2a = bq - 5b. Factoring out a and b, respectively, gives us a(x - 2) = b(q - 5). From here, let a = 0, and let q = 5. Now we have 0(x - 2) = b(5 - 5), which we can simplify to 0(x - 2) = b(0). Now we can see that x can be any value we want, but this will still result in 0 = 0. The same process can be done for q if we let b = 0 and x = 2.

Jonathan Messiers - 1 year, 1 month ago

Yeah but only if they r 0, this condition fails... else it works!!! Good problem

Balaji Subramoniam - 5 years, 5 months ago

Ahmed Obaiedallah
Dec 20, 2015

1 equation 4 independent variables If the number of variables > number of equations ... there will be either no solution or an infinite number of solutions in this case it's infinite solutions

I think the real problem is the distinct between the common understanding of the meaning of the term "implies" ("can be"), and the mathematical definition of the term ("can only be")

James Curran - 5 years, 5 months ago

I've thought the mathematical definition was: 'can be seen without rigorous inspection', which implies 'there could be other solutions, but we'll start here'.

Hazard Sharp - 5 years, 1 month ago

Exactly, because of this the correct answer must be true. What ever the a and b values are the above statement is true. Specifically for 0 value of a, there may be infinitely many solutions, though the above statement is again true - so 'can be'.

Oguz Tanatar - 5 years, 1 month ago

Richard Levine
Dec 23, 2015

ax + 5b = 2a + bq, so a(x-2) = b(q-5). Then a(x-2) - b(q-5) = 0. When a=0 and b=0, x and q can be any value.

Have you used any other values?

Mihail Belega - 4 years, 11 months ago

Wrong. That's in the case of multiplication when you can take either one of them zero

Sonia Khurana - 3 years, 5 months ago

Guillermo Templado
Dec 19, 2015

Let a=b and x = 7, q = 10, then LHS = ax + 5b = 7a + 5b = 12a and RHS = 2a + bq= 2a + b10 = 12a $\Rightarrow$ x = 7 is dintinct to 2, q = 10 is distinct to 5 and LHS = RHS . Therefore, the statement is false. LHS =RHS doesn't imply that x=2 and q=5

Krystal Brook
Dec 24, 2015

ax+5b=2a+bq can be re-written as: a(x-2) - b(q-5) = 0 As long as "a" OR "x-2" is 0 AND "b" or "q-5" is 0, the equation works.

Davy Ker
Dec 23, 2015

This is really an exercise in understanding basic logic, understanding what 'implies' means in mathematical analysis.

Enock Gita
Apr 25, 2020

The statement is false because if you let a=2,x=3,b=2 and q=6 L.H.S= $2\times3+2\times5=16$ ,R.H.S= $2\times2+2\times6=16$ so the equation can still hold without x and q being 2 and 5 respectively

Tim A
Oct 7, 2016

If a and b are made to equal each other then x = 2 and q = 5 could be true or x = 5 and q = 2 could also be true. In any case the answer is not static.

Is It True Because of Symmetry?

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