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.

Related questions
Impact of this question
1107 views around the world
You can reuse this answer
Creative Commons License