Simultaneous Equations Alternate Approach

[hackedagainanda]

Homework Statement

Figure out an easy way to solve this pair of equations.

(55x + 45y) = 520
(45x + 55y) = 480

Relevant Equations

(55x + 45y) = 520
(45x + 55y) = 480

(55x + 45y) = 520 (1)
(45x + 55y) = 480 (2)

So first I notice they are divisible by 5 so I go ahead and do that.

(11x + 9y) = 104 (1)
(9x + 11y) = 96 (2)

11 times 9 is 99 and 9 times 11 is 99 so I can cancel some terms. I proceed to do that by multiplying the top by 11 and getting: (121x + 99y) = 1144 and the bottom equation by 9 and I get (81x + 99y) = 864

I then have:
(121x + 99y) = 1144 (1)
(81x + 99y) = 864 (2)

So with the new equations I then subtract equation 2 from 1.

40x = 280, so x = 7.

Substituting x= 7 into the original equation 1 I get 55 x 7 = 385, so I subtract 385 from both sides of the equation and get 45y = 135 which is 135/45 = 3. y = 3

So I now have solved for x and y for both equations, and when I plug them in as values it gives me true equations.

This method seemed easy to me but I can't help wondering did I miss the point of the question, i.e is there a blindingly obvious easier method?

 

[FactChecker]

"blindingly obvious" -- I don't think so. But there are other approaches just as easy as yours.

 

[DaveE]

I think that's a great approach, for this particular problem. It may not be the easiest way in general, for another set of numbers.

That is why I favor methods that always work, even when the coefficients are obnoxious. The more general methods don't require you to be clever, you just have to follow the process, like a computer.

Cramer's rule is a good approach in general, but sometimes not the easiest. Perhaps you haven't studied that yet, but you will.

 

[robphy]

The arithmetic is much easier in this problem
if you take first the sum of the given equations, together with the difference of the given equations.

Of course, graphing the equations is another way.

 

[archaic]

If you add both equations, you get $100x+100y=1000$, which is equivalent to $x+y=10$.
$$55x+45y=520\Leftrightarrow11x+9y=104$$$$45x+55y=480\Leftrightarrow9x+11y=96$$$$x+y=10$$

 

[mathwonk]

or add and subtract, getting x+y = 10 and x-y = 4, so x=7, y =3.

 

[robphy]

I purposely didn’t finish carrying out the method I suggested above... hoping that OP would do it.

(The motivation of my suggestion (for me) was noticing the similarity with a Lorentz transformation, then using light cone coordinates.)

 

[hmmm27]

I got the "sum" bit right away ; totally missed "difference". Not any easier than the OP's method, but definitely cleverer.

 

[robphy]

I got the "sum" bit right away ; totally missed "difference". Not any easier than the OP's method, but definitely cleverer.

Really?
No scaling the way done by the OP was needed... since the numbers were nice.
Arithmetic simplifies quickly to what @mathwonk wrote.

 

[hmmm27]

Really?
No scaling the way done by the OP was needed... since the numbers were nice.
Arithmetic simplifies quickly to what @mathwonk wrote.

Yes, but you have to notice that both adding and subtracting the equations are useful, in the first place : the problem is rigged to have a cute answer.