xy=90 2x+y=36

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tiburtius | High School Teacher | (Level 2) Educator

Posted on

You need to solve system of equations

`{(xy=90),(2x+y=36):}`

From second equation we have

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

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

`x(36-2x)=90`

`36x-2x^2=90`

`-2x^2+36x-90=0`

`x^2-18x-45=0`

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

`y_1=36-2x_1=36-6=30`

`y_2=36-2x_2=36-30=6`

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

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tamtamtameena | Student, Grade 10 | (Level 1) eNoter

Posted on

Substitution Method

xy=90

2x+y=36

Rearrange the second formula to get y on its own

y=36-2x

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

x(36-2x)=90

36x-2x²=90

-2x²+36x=90

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

2x²-36x+90=0

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

x²-18x+45=0

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.

15y=90

y=90/15

y= 6

The second solution is (15,6)

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

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