2x - 3y = 2 --> 6x - 9y = 6

3x + 2y = 16 --> 6x + 4y = 32

Take the first from the second:

13y = 26 ∴ y = 2

Substitute in y = 2: 2x - (3*2) = 2, 2x = 8, x = 4

Check: 3*4 + 2*2 = 16.

**x = 4, y = 2**

You always use the method of making variables match when the two equations are linear. If they are quadratic, you need to use substitution.

The way to use substitution for the example above would be to rearrange the first equation to give x as the subject:

2x - 3y = 2

2x = 2 + 3y

x = (2 + 3y)/2

Then substitute that into the second equation.

3[(2 + 3y)/2] +2y = 16

That can be used to solve for y and then you would go back and use the y value to find x, as before.

3x + y = 8

y = 8 - 3x

My mistake was, I was always trying to take 3x over first.

]]>2x - 3y = 2

3x + 2y = 16

3y - (-10y) = 3y + 10y = 13y

]]>4x + 3y = 19

4x - 10y = 6

surely

0x + (3y - 10y) = 13

-7y = 13 and not 13y = 13 as you suggest. Your answer is indeed correct but I'm not quite sure how you got this part.

4x - 10y = 6

Pair this up with the other:

4x + 3y = 19

4x - 10y = 6

Subtract the second equation from the first: 13y = 13 ∴ y = 1

Use this to work out x: 4x - 10 = 6, 4x = 16, x = 4

Check with the other equation: (4*4) + 3*1 = 19. So, it does have whole numbers.

]]>4x + 3y = 19

2x - 5y = 3

I've tried getting each of the letters on their own and putting it back into the equations but for each, I end up with a fractional answer. The answers in the back of the book say they should be whole numbers!

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