To solve by elimination, you recognize that the coefficients of y in the two equations are opposites, so you can add the two equations to eliminate y.

.. (3x +3y) +(2x -3y) = (2) +(-17)

.. 5x = -15 . . . simplify

.. x = -3 . . . divide by 5

Now, you can substitute for x in either equation to find y.

.. 3(-3) +3y = 2

.. 3y = 11 . . . add 9

.. y = 11/3 = 3 2/3 . . . divide by 3

The solution is (x, y) = (-3, 3 2/3).

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One way to find points for graphing is to put the equations into intercept form. Divide by the constant on the right, and express each variable's coefficient as a denominator.

.. 3x/2 +3y/2 = 1

.. x/(2/3) +y/(2/3) = 1

This tells you that the x- and y-intercepts of the first equation are

.. (2/3, 0) and (0, 2/3)

Doing the same thing with the second equation tells you its x- and y-intercepts.

.. x/(-17/2) +y/(17/3) = 1

The intercepts for this equation are

.. (-8 1/2, 0) and (0, 5 2/3)

We usually like to graph points with integer coordinates. In the case of the first equation, that is not possible, as it can be rewritten as

.. x +y = 2/3

or

.. y = -x +2/3

So if one value is an integer, the other will not be.

The second equation can have integer points on its graph. We can recognize that when y=1, x will be an integer, so (-7, 1) is a point on the graph. Increasing x by 3 while y increases by 2 will give another point on the graph (-4, 3). Any odd value for y will give an integer value for x.