To solve systems of equations we want to eliminate (get rid of) one of the variables so we can solve for the remaining one, then use our solution to find the value of the second variable.
The tricky part is keeping track of the signs, so observe the rules carefully. Do not change signs until you must.
Given: #-6x -4y = -10# for equation #(1)#
And: #-8x -9y = 16# for equation #(2)#
To eliminate the #x#'s we can multiply the entire equation #(1)# by the coefficient or #x# multiplier in equation #(2)# and then multiply the entire equation #(2)# by the #x# multiplier from #(1)#.
#WHAT???# See below:
#(1) rArr (-6x -4y = -10) xx 8# from #(2)#
#(2) rArr (-8x -9y = 16) xx 6# from #(1)#
#-48x -32y = -80#
#-48x -54y = 96#
Now we want to subtract #(2)# from #(1)#, so we will need to change all the signs in #(2)#; since in order to subtract, the parts you are taking away become negative, then you can simply add.
#-48x -32y = -80#
#48x +54y = -96#
Then: #22y = -176#
#y = -176/22 = -8#
Solve for #x# by substituting your value for #y# into the original equation #(1)#:
#-6x -4y = -10#
#-6x -4(-8) = -10#
#-6x = -32 - 10#
#-6x = -42#
#x = (-42)/-6#
#x = 7#
Always check:
Check your work by substituting your value for #x# and #y# into the original equation #(2)#:
#-8x -9y = 16#
#-8(7) -9(-8) =16#
#-56 + 72 = 16#
#16 = 16#
There are other ways of doing this, but if you can follow this method it will be useful in solving bigger systems in the future.
