Help with solving problems using two variables and equations

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Here is the problem i need help with:

Mrs. Mabunga deposited Php 18,000 in two separate account -- a certain amount at 6% annual interest, and the remaining amount at 7%. If at the end of the year, the interest gained from both accounts was Php 1190, how much did she invest at each rate?

I need help in answering the problem using TWO variables and TWO equations. THANK YOU! Also, please explain WELL on how you got the answer.

Let x be the amount invested at 6% and y be the amount invested at 7%. Then the first fact translates as:

x + y = 18,000

The one-year interest amount at 6% on an investment x is 0.06x. Similarly the 7% account earns 0.07y in the same year. That means the second fact translates to:

0.06x + 0.07y = 1,190
6x + 7y = 119,000 ... multiplying by 100 to remove fractions.

Now you have two equations in two unknowns to solve:

x + y = 18,000 .... equation 1
6x + 7y = 119,000 .... equation 2

Multiply both sides of eq. 1 by 6:

6x + 6y = 108,000 .... equation 3

Subtract equation 3 from equation 2:

6x + 7y - 6x - 6y = 119,000 - 108,000
y = 11,000

That's one part of the answer. Plug this into equation 1 to get:

x + 11,000 = 18,000
x = 7,000

Then test: 7,000*0.06 + 11,000*0.07 = 420 + 770 = 1,190

Let x = the amount she invested at 6%
Let y = the amount she invested at 7%

The total of the two amounts was 18000

x + y = 18000

6% of the amount she invested at 6%, plus 7% of the amount she invested at 7%, will be equal to 1190.

0.06x + 0.07y = 1190

So you have your two equations

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Multiply the second equation by 100 to clear the decimals.

x + y = 18000
6x + 7y = 119000

Multiply the first equation by -6

-6x - 6y = -108000
6x + 7y = 119000