Show that the system of equations below has a unique solution if and only if bc≠ad

Question

bc≠ad

ax+by=s

cx+dy=t

Hint: Eliminate either x or y by using the addition method

Answer 1

ax + by = s

cx + dy = t

x + (b/a)y = s/a --- divide by a, if a != 0 Equation A

x + (d/c)y = t/c --- divide by c, if c != 0 Equation B

(b/a - d/c)y = s/a - t/c --- subtract Equation B from Equation A

y = (s/a - t/c) / (b/a - d/c) --- divide both sides by (b/a - d/c), if (b/a - d/c) != 0

but if (b/a - d/c) != 0

==> (bc-ad) / ac != 0 --- create a common denominator, 'ac'

==> (bc - ad) != 0 --- multiply both sides by 'ac'

==> bc != ad --- add 'ad' to both sides

So we have shown that solving the system of equations requires that 'a' and 'c' are not zero, and 'bc' != 'ad'.

Notation: '==>' is 'implies', and '!=' is ≠.

Answer 2

Hi, Jane.

Okay, so you have:

ax+by = s

cx + dy = t

You need to show that in order to have a unique solution, bc ≠ ad.

So, let's eliminate the x:

First, multiply the first equation by c:

c(ax + by = s) = acx + bcy = cs

Next, multliply the 2nd equation by a:

a(cx + dy = t) = acx + ady = at

Now, subtract the new equations to eliminate the x variable:

acx + bcy = cs

-acx - ady = -at

-------------------------

bcy - ady = cs - at

(bc - ad)y = cs - at

However, if bc = ad, then you would get :

0y = cs - at , which would not give a unique solution. Therefore, In order to have a unique solution,

bc ≠ ad.