# 10x+10y+2xy=278 and 5xy=240. Solve with substitution and solve with quadratic equation.

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Solve the system 10x+10y+2xy=278;5xy=240:

First, solve the second equation for y in terms of x. (You could also solve for x in terms of y.)

`5xy=240 ==> y=48/x` Now substitute for y in the first equation:

`10x+480/x+(2x)(48/x)=278`

`10x+480/x+96=278`

`10x+480/x=182` Multiply both sides by x:

`10x^2+480=182x` Write in standard form and solve:

`10x^2-182x+480=0`

This factors as 2(5x-16)(x-15)=0 or:

`2(5x^2-91x+240)=0` Using the quadratic formula:

`x=(91+-sqrt(91^2-4(5)(240)))/(10)`

`x=(91+-sqrt(3481))/10`

`x=(91-59)/10 "or" x=(91+59)/10`

` `x=3.2 or x=15

If x=3.2 then `y=48/3.2=15`

If x=15 then `y=48/15=3.2`

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**The solutions are (3.2,15) and (15,3.2)**

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The graph of 10x+10y+2xy=278 in red and the graph of 5xy=240 in black:

We can simplify the 10x + 10y + 2xy = 278, by dividing each sides by 2.

So, 5x + 5y + xy = 139.

We can divide both sides of 5xy = 240 by 5. So, it will bcomes xy = 48.

Hence, we can replace the xy on 5x + 5y + xy = 139 by 48.

So, 5x + 5y + 48 = 139.

Subtract both sides by 48. 5x + 5y = 91.

We can solve for x using xy = 48. So, x = 48/y.

We can use this to solve for y on 5x + 5y = 91.

Plug-in x = 48/y.

5(48/y) + 5y = 91

Multiplying both sides by y:

240 + 5y^2 = 91y.

Subtract 91y on both sides.

5y^2 - 91y + 240 = 0.

Factoring the left side:

(y - 5)(5y - 16) = 0

Equaing each factor to zero. And solving for x.

y = 15, y = 16/5

Let solve for x using those values of y.

x(15) =48 , x = 48/15.

Simplifying 48/15, by dividing the top and bottom by 3. x = 16/5.

Solving for x, when y = 16/5.

x = 48/(16/5) = 48(5/16) = 3(5) = 15.

Hence, solutions here are **(15, 16/5) and (16/5, 15)**.

10x+10y+2xy=278 (i)

5xy=240 (ii)

substutute xy from (ii) in (i)

we have

5x+5y=91 and xy= 48

x+y=91/5 and xy=48

we know

(x-y)^2=(x+y)^2-4xy

(x-y)^2=(91/2)^2-4 . 48=331.24-192=139.24

x-y=11.8 or x-y= -11.8 (iv)Thus we have now

x+y=18.2 and x-y=11.8

or

x+y=18.2 and x-y= -11.8

solving these system of equations we have

x=15,3.2

y=3.2,15 i.e either (15,3.2) or (3.2,15)

Ans.