Solve the system of linear equations below. 6x + 3y = 33 4x + y = 15 Select one of the options below as your answer: A. x = 2, y = 7 B. x = -13, y = 7 C. x = - 2 3 , y = 12 2 3 D. x = 5, y = 1
are you familiar with the elimination method ?
somewhat
multiply the 2nd equation by -3
or...would you rather do substitution ?
substitution is far easier for me
ok....look at the 2nd equation...can you isolate y ?
BY adding 4x to the opposite side?
subtracting*
no...subtract 4x from both sides
so y=15-4x
correct....now sub 15 - 4x in for y in the 1st equation
6x+3(15-4x)=33?
yes......now distribute the 3 through the parenthesis
6x+45-12x=33. Now C.L.T right?
correct....combine like terms
-6x+45=33
correct....now subtract 45 from both sides...or add 6x to both sides...whichever you want to do
but if you add 6x, you also have to subtract 33 from both sides
-6x=-12 now divide by negative 6?
correct....now divide both sides by 6 keeping in mind dividing by 2 negatives gives you a positive
so that gives me x=2 now I put that in for x on one of the equations right
correct
and theen solve the same way
yes
you can then check your answers by subbing both x and y into either of the equations and if they come out equal, you have found the correct answers
ok I put x into the first equation listed and then have gotten to 12=33-3y. What now?
6(2) + 3y = 33 12 + 3y = 33 how did you get that ? from here you subtract 12 from both sides
ohhhh I subtracted 3y first
you are trying to get 3y on one side and everything else on the other side
you got it now...good job
Join our real-time social learning platform and learn together with your friends!