Help with provinga simple algebraic equation?

[cshum00]

Homework Statement

ar_1 = b <=> r_1 = \frac{b}{a} (equation1)
cr_d = c <=> r_2 = \frac{d}{c} (equation2)
ar_1 + cr_2 = b + c <=> r_1 + r_2 = \frac{b}{a} + \frac{d}{c} (add equation1 and equation2)

The Attempt at a Solution

I am trying to make the equation on the left equals to the equation on the right side. I have tried adding, subtracting and dividing values, also multiplying and dividing conjugates but nothing works.

 

[Дьявол]

I do not understand something with the second equation.

Should it be

<br /> cr_2 = d &lt;=&gt; r_2 = \frac{d}{c}<br />
?

What do you need to prove? What is the original text of the task?

 

[cshum00]

I have 2 equations.
ar_1 = b (equation1)
and cr_d = c (equation2)

i just divided by 'a' in both sides in equation1
and divided by 'c' in both sides in equation2
so i get
r_1 = \frac{b}{a} (equation1)
r_2 = \frac{d}{c} (equation2)

but they are both reversible so
ar_1 = b &lt;=&gt; r_1 = \frac{b}{a} (equation1)
cr_d = c &lt;=&gt; r_2 = \frac{d}{c} (equation2)

if i add both equations i get:
ar_1 + cr_2 = b + c &lt;=&gt; r_1 + r_2 = \frac{b}{a} + \frac{d}{c}
since both of the equations are the same, so should the sum be
and that is what i am trying to prove.

Any numerical examples i try, both expressions seem to be the same except for 'a' and 'c' equals to zero.

 

[HallsofIvy]

Homework Statement

ar_1 = b &lt;=&gt; r_1 = \frac{b}{a} (equation1)
cr_d = c &lt;=&gt; r_2 = \frac{d}{c} (equation2)
ar_1 + cr_2 = b + c &lt;=&gt; r_1 + r_2 = \frac{b}{a} + \frac{d}{c} (add equation1 and equation2)

The Attempt at a Solution

I am trying to make the equation on the left equals to the equation on the right side. I have tried adding, subtracting and dividing values, also multiplying and dividing conjugates but nothing works.

You can't prove it, it isn't true. Knowing that ar_1+ br_2= b+ c does NOT tell you that ar_1= b and ar_2= c separately.

Knowing the two things ar_1= b and cr_d= c tells you, by adding the equations, that ar_1+ cr_d= b+ c but knowing that single fact, that ar_1+ cr_d= b+ c does NOT tell you the two things.

For example, If x= 3 and y= 2 then x+ y= 5. But knowing only that x+y= 5 does NOT tell you that x= 3 and y= 2.

 

[cshum00]

Then what about this case?

The parametric equation of a ellipse is:
x = acos \theta
y = bsin \theta

I first square both equations so I get:
x^2 = a^2cos^2 \theta (1)
y^2 = b^2sin^2 \theta (2)

Then for equation (1) divide by a^2 and equation (2) divide by b^2:
\frac{x^2}{a^2} = cos^2 \theta (3)
\frac{y^2}{b^2} = sin^2 \theta (4)

So if i were to add equations (1) and (2) i would get:
x^2 + y^2 = a^2cos^2 \theta + b^2sin^2 \theta (5)
And adding equation (3) and (4)
\frac{x^2}{a^2} + \frac{y^2}{b^2} = cos^2 \theta + sin^2 \theta (6)

Then putting them in my format i would get:
x^2 + y^2 = a^2cos^2 \theta + b^2sin^2 \theta &lt;=&gt;
\frac{x^2}{a^2} + \frac{y^2}{b^2} = cos^2 \theta + sin^2 \theta

Which both are equations of the ellipse the only difference is that one I can remove
the third variable of \theta by using the trigonometry identity of sin^2\theta + cos^2\theta=1.

 

[Mark44]

Then what about this case?

The parametric equation of a ellipse is:
x = acos \theta
y = bsin \theta

I first square both equations so I get:
x^2 = a^2cos^2 \theta (1)
y^2 = b^2sin^2 \theta (2)

Then for equation (1) divide by a^2 and equation (2) divide by b^2:
\frac{x^2}{a^2} = cos^2 \theta (3)
\frac{y^2}{b^2} = sin^2 \theta (4)

So if i were to add equations (1) and (2) i would get:
x^2 + y^2 = a^2cos^2 \theta + b^2sin^2 \theta (5)
And adding equation (3) and (4)
\frac{x^2}{a^2} + \frac{y^2}{b^2} = cos^2 \theta + sin^2 \theta (6)

Then putting them in my format i would get:
x^2 + y^2 = a^2cos^2 \theta + b^2sin^2 \theta &lt;=&gt;
\frac{x^2}{a^2} + \frac{y^2}{b^2} = cos^2 \theta + sin^2 \theta

Which both are equations of the ellipse the only difference is that one I can remove
the third variable of \theta by using the trigonometry identity of sin^2\theta + cos^2\theta=1.

It's fairly simple to go from the parametric equations for an ellipse
x = acos \theta
y = bsin \theta

to the Cartesian form:
x^2/a^2 + y^2/b^2 = 1
You almost got to this form using way too many steps.

Your original equations were garbled and, I believe, confused several people responding. As far as I can tell, your equations should have been:
ar_1 = b
cr_2 = d
ar_1 + cr_2 = b + d
(For your second equation you had cr_d = c which I'm nearly certain is an error. Even though Дьявол asked you whether this was an oversight, you didn't respond to him and persisted with the erroneous equation.)
Your 3rd equation essentially says that b + d = b + d, which is trivially true. There is no way to divide the first term in the third equation by a and the second term by c. IOW, you can't transform your third equation to r_1 + r_2 = &lt;whatever&gt;

 

[cshum00]

Ya the second line was an error. I should have been cr_2 = d &lt;=&gt; r_2 = \frac{d}{c}.

Sorry Дьявол. I did not notice that. And because i didn't notice, i was confused by his question of what needed to be proven; also copied the equations wrong for the second time.

Ya, the ellipse it just an simple example. I tried to write it in general form at the beginning. So going back to the general examples, if you actually try to plug some numbers you see that in the third equation both sides are the same except for "a" and "c" = 0.