The power of these equations is best demonstrated by working through several examples.
If ba=316, what is a−ba+b?
By using equation 3 above with k=1, we know that if ba=dc, then a−ba+b=c−dc+d. Thus,
Without using the theorem, one approach would be to substitutea=316b to obtain that the expression is equal to 316b−b316b+b. This can be eventually simplified to the answer of 1319 but will involve more steps.
If a−ba+b=1319, what is ba?
We will apply equation 3 again to obtain that
Simplifying both sides, we obtain 2b2a=632⟹ba=316. □
Observe that this answer matches the condition given in the previous example. This is not surprising, as the statements are converses of each other. In fact, the converse of componendo can be proved by applying the theorem again as done above.
If ba=dc, prove that 2a+9b2a−9b=2c+9d2c−9d.
ba9b2a2a−9b2a+9b2a+9b2a−9b=dc=9d2c=2c−9d2c+9d=2c+9d2c−9d.□(mutliplying both sides by 2 and dividing both sides by 9)(by componendo and divdendo)(by invertendo)
Method 2. Alternatively, we can make use of the property that a−kba+kb=c−kdc+kd:
If ba=dc, then
Multiplying both the numerator and the denominator by 2 on both sides, we get
If 3a2b+b3a3+3ab2=3x2y+y3x3+3xy2, prove that ax=by.
3a2b+b3a3+3ab2a3+3ab2−3a2b−b3a3+3ab2+3a2b+b3a3−b3−3ab(a−b)a3+b3+3ab(a+b)(a−b)3(a+b)3(a−b)(a+b)(a+b)−(a+b)(a+b)+(a−b)2b2abaax=3x2y+y3x3+3xy2=x3+3xy2−3x2y−y3x3+3xy2+3x2y+y3=x3−y3−3xy(x−y)x3+y3+3xy(x+y)=(x−y)3(x+y)3=(x−y)(x+y)=(x+y)−(x+y)(x+y)+(x−y)=2y2x=yx=by.□(by componendo and dividendo)(taking cube root on both sides)(by alternendo)
What quantity must be added to each term of the ratio a−2ba+2b, to make it equal to (a−b)2(a+b)2?
3x−5y3x−2xxyx5y3x3x−5y3x+5y=2x+y=y+5y=6y=16=5×13×6=518=18−518+5=1323.□(mutliplying both sides by 3 and dividing both sides by 5)(by componendo and dividendo)
Solve for x: x−5x+5=2.
Applying equation 3 above with k=1, we obtain
This simplifies to
If ba=dc, show that b+naa+mb=d+ncc+md for n=a−b.
Method 1: Direct Proof
Method 2: Using Theorem 4
bacac+mda+mbb+naa+mb=dc=db=d+ncb+na=d+ncc+md.(by invertendo)(by theorem 4)(by invertendo again)
Note that Method 2 does require the additional condition m=d−c, though if m=d−c the statement is trivially true. □
If the ratio of x+y to x−y is 411, then the ratio of y to x can be written as ba, where a and b are coprime positive integers. Find a+b.
Thus far, the examples have only involved linear terms for simplicity. The variables a and b are allowed to be polynomials or even exponential functions. Consider the following example, where the fraction isn't initially in the form of a−ba+b:
Solve for x: x+1x3+1=x−1x3−1.
For the fractions to be defined, we must have x=1,−1.
While the fractions are not in a form that allows an immediate application of componendo-dividendo, we can cross multiply the denominators to obtain
This is in a recognizable form, so we apply Componendo and Dividendo with k=1(which is valid since x−1x+1=1). Then we get that
However, since x=1,−1, we have x=0 as the only solution. □
and thus x(xn−1−1)=0. We need to exclude both x=1 and x=−1 as solutions. Thus,
there are (n−1)roots when n is even, namely 0 and the (n−2) complex (n−1)throots of unity(1 omitted);
there are (n−2) roots when n is odd, namely 0 and the (n−3) complex (n−1)th roots of unity (1 and −1 omitted).
If n=1, then there are infinitely many roots.
If n=0, then there are no solutions.
If n<0, then the same algebraic steps as above hold. But we have to exclude 0 as a root, so the roots are the ∣n−1∣th roots of unity, with 1 and −1 excluded. □
If tanθ=−41, find the value of cosθ−3sinθsinθ+2cosθ.
Using b+naa+mb=d+ncc+md (see the proof in an above example) and rewriting the given condition as cosθsinθ=4−1, we immediately get
Note that we could have also written the given condition as cosθsinθ=−41, but this would have not changed the final answer. □