Question

Solve the following system of equations using the substitution method. Enter your answers as (x,y) pairs, separated by commas.
x+3y=14
xy=11

Answer 1

x= 14-3y
(14-3y)y = 11
14y-3y^2 - 11 =0
3y^2-14y+11=0
3.6666666 , 1
x1 = 14-3*3.66666 = 3
x2 = 14 - 3 = 11
(3,3.6666) , (11,1)

notice that 3.66666 - is 3 and 2 thirds

Answer 2

I would solve the top equation for x
x = 14-3y
then substitute into the bottom
(14-3y)y=11
14y-3y^2=11
Get all the terms together
-3y^2+14y-11=0
Use quadratic formula to find y (you will get two values)
Use those values to find the x values

Answer 3

I already posted the solution for this. This question has been posted twice.

Answer 4

That seems to be happening tonight a bunch

Answer 5

1. Solve both equations for y:

y = (14-x)/3

y = 11/x

Set y = y:

(14-x)/3 = 11/x

Cross Multiply:

x(14-x) = 3*11

14x - x^2 = 33

Put in Standard form:

x^2 -14x + 33 = 0

Solve for x:

x^2 - 11x - 3x + 33 = 0

x(x-11) - 3(x-11) = 0

(x-11)(x-3) = 0

x = 3, 11

y = 1, 11/3

(11,1)
(3,11/3)

Answer 6

Thank you all