How do you solve this system of equations: #x+ 3y = - 3 and 4x + 11y = - 14#?
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Let's choose a variable to cancel out. For this one, I'll choose #x#. You find the LCM of the coefficients of that variable (you want the same amount of #x# in each equation so that you can subtract and get rid of it). The LCM of #1# and #4# is #4#, so I'll multiply the first equation by #4# on both sides (if I only did one side, the equation wouldn't be equal anymore).
#4*(x+3y)=4*(-3)#
#4x+12y=-12#
Now, I can subtract the second equation from the first equation (or the first from the second. Either works.)
#(4x+12y)-(4x+11y)=(-12)-(-14)#
#4x+12y-4x-11y=-12+14#
#4x-4x+12y-11y=2#
#y=2#
Knowing this, I can plug into any equation to get the value of #x#.
#x+3(2)=-3#
#x+6=-3#
#x=-3-6#
#x=-9#
We can check these values in our other equation just be sure they're correct:
#4(-9)+11(2)=-14#
#-36+22=-14#
#-14=-14#
Since we got what we were looking for, our values are correct.