Expert Answers
tiburtius eNotes educator| Certified Educator

You need to solve system of equations


From second equation we have

`y=36-2x`.                                                            (1)

So we put that into first equation instead of `y`.





We can wow solve this quadratic equation by using this formula` `

`x_(1,2)=(-b pm sqrt(b^2-4ac))/(2a)`

where `a, b` and `c` are coefficients of equation `ax^2+bx+c=0`.

So in your case

`x_(1,2)=(18pm sqrt(324-180) )/2=(18pm12)/2`

`x_1=3`, `x_2=15`

Now by putting this into (1) we get



So your solutions are `(x_1,y_1)=(3,30)` and `(x_2,y_2)=(15,6)`.

tamtamtameena | Student

Substitution Method



Rearrange the second formula to get y on its own


Substitute 'y=36-2x' into the first equation




Move the -2x² and +36x to the other side to turn the -2x² into positive 2x²


Divide the whole equation by 2 to change 2x² to just x² as its easier to deal with.


Factorise this quadratic into brackets.

Two numbers that times together to make 45 and add to make -18 are -3 and -15

(x-3) (x-15)=0 

x= 3 x= 15 

Substitute 3 for x to get 3y=90 

Rearrange it to get y= 90/3

y= 30

The first solution is (3,30)

Now for the second solution substitute 15 into the the equation.



y= 6

The second solution is (15,6)

The two solutions are (3,30) and (15,6)