Kenneth H. answered • 05/17/19
Stanford Physics Grad and Experienced Tutor for any grade level
For solving systems of equations, you can always use substitution (in which you get x by itself on one side of one of the equations and then take what's on the other side and substitute it for x in the other equation) but in the case of linear equations like you've got here, it's easier to cross multiply and add the equations to eliminate one of the variables. Here's what the steps for that look like:
-2x + 3y = -6
5x - 6y = 15
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5*(-2x + 3y) = 5*(-6)
2*(5x - 6y) = 2*(15)
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-10x + 15y = -30
10x - 12y = 30
Then we just add the two equations straight down, combining like terms, so the -10x in the first equation adds with the 10x in the second, cancelling out the x's. Similarly adding the other terms together, we're left with
0 + 3y = 0
Or just y = 0, and that's your final answer. If you wanted to find the corresponding x coordinate, you would just plug 0 in for y in either one of the equations.
(One quick note, when you use this method for other problems, notice that you want one of the variables to get cancelled out, so you multiply either equation by whatever numbers will result in that variable getting cancelled, which means one term in one equation has to end up being the negative of the same term in the other equation. So if we wanted to cancel out the y's, we could just multiply the first equation by 2, so you get 6y and -6y. Remember that whenever you use this method, you have to multiply the whole equation on both sides by whatever number you've picked to cancel out a variable.)
