How do you solve this system of equations: #x+ 3y = - 3 and 4x + 11y = - 14#?

1 Answer
YoshiMan007
Oct 15, 2017

#x=-9, y=2#

Explanation:

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.

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