There are basically 2 types of 2-variable equations you will be asked to solve. The first type is asking, what is y when x is _____? For example, for the linear equation y = 3x + 4....what is y when x is 2? Just plug in 2 for x and solve for y...... y = 3(2) + 4. When x is 2, y is 10. That solution looks like: (2, 10). In linear equations, there is a different y for every x...just plug it in.

The second type of 2 variable problem solving is when you have two (or more) equations with the same solution; the point where the equations meet on the graph. This is called "systems of equations" (and can have more than 2 variables, but that's a future discussion). For example: (stacking them makes them easier to solve) (often they are in "standard" form versus "slope intercept" (y = mx + b) form)

2x - 4y = 2

3x + 5y = 14

In math, there are OFTEN many ways to solve a problem, choose the easiest way.. the more ways you know how, the more choices you have!

In this case, the easiest way is to "eliminate one of the variables" by changing one or both of the equations into an equal form through multiplication. OK, I could either multiply the top equation by - 3 and the bottom equation by 2 to eliminate the x:

(-3) 2 x - 4 y = 2 changes into -6 x + 12 y = -6

(2) 3 x + 5 y = 14 changes into 6 x + 10 y = 28 (see how the 6 x cancels the -6x?)

When we combine the new equations, we get a nice 1 variable result: 22 y = 22 (so y = 1 !)

Now, I can plug that y into either equation, 2x - 4 (1) = 2 2x-4 = 2 2 x = 6. (so x = 3)

Our solution is (3, 1). Try plugging those values back into both equations to satisfy you're correct! What that means is when you draw those two linear equations on the graph, they will INTERSECT at the point (3 , 1). that's the solution!

Can you solve the equation another way? Try multiplying the top equation by 5 and the botton equation by 4, which will eliminate the y,

10 x - 20 y = 10

12 x + 20 y = 56

so solve for x and continue the same way as above and check your solution!