A System of Nonlinear Equations

Algebra Level 2

Given the system of equations { x ( x + y ) = 9 y ( x + y ) = 16 \begin{cases} x(x+y) &=& 9 \\ y(x+y) &=& 16 \end{cases} the value of x y xy can be written as a b \frac{a}{b} where a a and b b are positive coprime integers. Find a + b a+b .


The answer is 169.

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12 solutions

Priyansh Sangule
Oct 28, 2013

Okay.

First we add the two equations to find out ( x + y ) (x+y) .

x ( x + y ) + y ( x + y ) = 25 x(x+y) + y(x+y) = 25

( x + y ) 2 = 5 2 \Rightarrow (x+y)^2 = 5^2

Now to get x y xy we multiply the two equations :

x y ( x + y ) 2 = 9 16 xy \cdot (x+y)^2 = 9 \cdot 16

x y = 144 25 \Rightarrow xy = \dfrac{144}{25}

Which is of the form a b \dfrac{a}{b}

Now since , a a and b b are coprime ,

a + b = 144 + 25 = 169 a + b = 144 + 25 = \boxed{169}

Therefore , 169 169 is the correct answer .

Cheers!

For those who didn't get the third line -

x ( x + y ) + y ( x + y ) = 25 x(x+y) + y(x+y) = 25

Here , we simply common out ( x + y ) (x+y) to get -

( x + y ) ( x + y ) = 5 2 (x+y) \cdot (x+y) = 5^2

Tried my best to make it understandable , elegant and short .

Priyansh Sangule - 7 years, 7 months ago

( x + y ) (x+y) can be 5 or -5.

Snehdeep Arora - 7 years, 7 months ago

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True , I completely forgot to include that point too.

Putting : ( x + y ) 2 = 5 2 (x+y)^2 = 5^2

Will not change the solution.

BTW , the value of :

( x + y ) = ± 5 (x+y) = \pm 5

Priyansh Sangule - 7 years, 7 months ago

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yes it won't change the solution but both the equations are satisfied when x and y are of the same sign i.e either positive or negative

Snehdeep Arora - 7 years, 7 months ago

In fact, there isn't a need to even say that ( x + y ) = ± 5 (x+y) = \pm 5 . I have removed that distracting detail from your solution, which makes it much better.

This was the approach that I used to solve (and create) this problem.

Calvin Lin Staff - 7 years, 7 months ago

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@Calvin Lin Thank you very much for helping me sir . I intended to do the same ( If I had the opportunity)

Thanks a lot.

Priyansh Sangule - 7 years, 7 months ago

Awesome dude! Exactly the way I did it! I mean ... EXACTLY!

David Kroell - 7 years, 7 months ago

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That's cool coincidence :D

Priyansh Sangule - 7 years, 7 months ago

xy(x+y)^2 = 9*16/25

Debjyoti Chattopadhyay - 7 years, 6 months ago
Justin Wong
Oct 27, 2013

Adding the equations after distribution,

x 2 + x y = 9 x^2+xy=9 plus

+ x y + y 2 = 16 +xy+y^2=16 equals

x 2 + 2 x y + y 2 = 25 x^2+2xy+y^2=25

Recognizing the perfect squares,

( x + y ) 2 = 5 2 (x+y)^2=5^2 .

Taking square roots,

x + y = 5 x+y=5 .

Substituting,

x ( 5 ) = 9 x(5)=9 and y ( 5 ) = 16 y(5)=16 .

Solving for variables,

x = 9 5 x=\frac{9}{5} and y = 16 5 y=\frac{16}{5} .

Multiplying for the answer x y xy ,

144 25 \frac{144}{25} means a + b = 169 a+b=169 .

Why must we have x + y = 5 x+y = 5 ?

Calvin Lin Staff - 7 years, 7 months ago

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I think I see now, Calvin.

It is false to say that x + y = 5 x+y=5 , but it is possible to prove that ( x + y ) 2 = 25 (x+y)^2=25 . More specifically, x + y x+y has two values and so a statement assuming only one of these values would be incorrect. That means my solution was a bit incorrect and I see the value of Priyansh's solution in that he just avoided the whole mess. In answer to your question, "We mustn't." If I really wanted to continue the method I used, I should make two cases for x + y = 5 x+y=5 and x + y = 5 x+y=-5 , solve each, then show they are equivalent. But that is more work.

Silly me... Thanks for helping me be a better mathematician. I appreciate how you always address flaws in solutions.

Justin Wong - 7 years, 7 months ago

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This question was designed to catch such a mistake, namely making a claim that x 2 = 1 x = 1 x^2 =1 \Rightarrow x = 1 . Of course, when stated this way, it seems obvious to everyone. I just disguised it in the form ( x + y ) 2 = 25 ( x + y ) = 5 (x+y)^2 = 25 \Rightarrow (x+y) = 5 .

Good identification of the error and reflecting on how to resolve the issue. Keep u the good work, and I'm sure you'd soar.

Calvin Lin Staff - 7 years, 7 months ago

I know that you're right but I don't understand how you get rid of the 2xy in the third step to give the (x+y)^2 = 25?

Ella Dunlop - 7 years, 7 months ago

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Because ( x + y ) 2 (x + y)^{2} = x 2 x^{2} + 2xy + y 2 y^{2}

Rindell Mabunga - 7 years, 7 months ago

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thanks!

Ella Dunlop - 7 years, 7 months ago
Prasad Nikam
Jan 13, 2014

X(X+Y)=9.....EQN.1 Y(X+y)=16....EQN.2 Dividing 1 by 2. We get:X=9Y Now substracting 1 and 2 we get: X(X+Y)-Y (X+Y)=-7 Y.Y-X.X =7 By Substituting: we get: Y=16/5 and X=9/5 Then XY=144/25 X+Y=144+25=169

Rindell Mabunga
Oct 28, 2013

If we let

x ( x + y ) = 9 be (eq 1)

and

y ( x + y ) = 16 be (eq 2)

then (1) + (2):

x ( x + y ) + y ( x + y ) = 9 + 16 = 25

by factoring:

( x + y ) × ( x + y ) ( x + y ) \times ( x + y ) = 5 × 5 5 \times 5

or

( x + y ) 2 ( x + y )^{2} = 5 2 5^{2}

taking the square root of both sides, we will get

x + y = 5 (eq 3) or x + y = -5 (eq 4)

Substitute (eq 3) and (eq 4) to (eq 1) and (eq 2) and simlifying, we can get

x = 9 5 \frac{9}{5} or 9 5 \frac{-9}{5}

y = 16 5 \frac{16}{5} or 16 5 \frac{-16}{5}

Therefore

xy = 9 5 × 16 5 \frac{9}{5} \times \frac{16}{5} or 9 5 × 16 5 \frac{-9}{5} \times \frac{-16}{5}

which both yields

xy = 144 25 \frac{144}{25}

Therefore, a = 144 and b = 25 and a + b = 169

Mohith Manohara
Oct 27, 2013

x 2 + x y = 9 x^{2}+xy=9

y 2 + x y = 16 y^{2}+xy=16

Adding the two equations give us this:

x 2 + 2 x y + y 2 = 25 x^{2}+2xy+y^{2}=25

( x + y ) 2 = 25 (x+y)^{2}=25

x + y = 5 x+y=5

Substituting into the original, we get:

5 x = 9 5x=9

x = 9 5 x=\frac{9}{5}

5 y = 16 5y=16

y = 16 5 y=\frac{16}{5}

Multiplying it together, we get 16 5 × 9 5 = 144 25 \frac{16}{5}\times\frac{9}{5}=\frac{144}{25}

144 + 25 = 169 144+25=\boxed{169}

Why must we have x + y = 5 x + y = 5 ?

Calvin Lin Staff - 7 years, 7 months ago

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I forgot to do something. ( x + y ) 2 = 25 (x+y)^{2}=25 , so x + y = ± 5 x+y=\pm{5} . It will give you the same answer regardless.

Mohith Manohara - 7 years, 7 months ago
Ahmad Awalluddin
Feb 5, 2014

we note that x+y=a

so ax=9 and ay=19

ax+ay=25 then we can simplified them to a^2=25.

we multiply the first term and second term which we will got xya^2=144

OK now is the last step to find the answer by replace a^2 with 25

so a+b =169

Andre Yudhistika
Jan 5, 2014

sum all of the equation and you'll get x^2+2xy+y^2=25

(x+y)^2=25 >>> x+y=5

therefore 5x=9 >>>x=9/5

5y=16 >>>> y=16/5

xy=9/5*16/5=144/25=a/b

a+b= 144+25=169

Adya Jaiswal
Nov 3, 2013

On adding both equations, we get -

x 2 + 2 x y + y 2 = 25 x + y = 5 x^2 + 2xy + y^2 = 25 \Rightarrow x+y=5

On subtracting e q n ( i i ) eqn(ii) from e q n ( i ) eqn(i) , we get-

y 2 x 2 = 7 ( y + x ) ( y x ) = 7 y x = 7 5 y^2-x^2=7 \Rightarrow (y+x)(y-x)=7 \Rightarrow y-x=\frac{7}{5}

This gives x = 9 5 x=\frac{9}{5} and y = 16 5 y=\frac{16}{5} . Hence x y = 144 25 = a b xy=\frac{144}{25}=\frac{a}{b}

Therefore, a + b = 169 a+b=169

Dividing the two equations, we get $$\frac{x}{y}=\frac{9}{16} \Rightarrow \frac{y}{x}=\frac{16}{9}.$$ Expanding the equations and adding, we get $$x^2+y^2+2xy=25.$$ Because x 2 + y 2 x^2+y^2 can be expressed as x y ( x y + x y ) xy(\frac{x}{y}+\frac{x}{y}) , we can rewrite our summed equation as $$xy(\frac{x}{y}+\frac{x}{y})+2xy=25.$$ But x y ( x y + x y ) = x y ( 9 16 + 16 9 ) = 337 x y 144 xy(\frac{x}{y}+\frac{x}{y})=xy(\frac{9}{16}+\frac{16}{9})=\frac{337xy}{144} . Now our summed equation becomes $$\frac{337xy}{144}+2xy=25 \Rightarrow \frac{625xy}{144}=25 \Rightarrow xy=\frac{144}{25}=\frac{a}{b}.$$ Therefore, a + b = 144 + 25 = 169 . a+b=144+25=\boxed{169}.

I meant to put $$xy(\frac{x}{y}+\frac{y}{x})$$ instead of $$xy(\frac{x}{y}+\frac{x}{y})$$.

Garrett Higginbotham - 7 years, 7 months ago
Hema Lekhaa
Oct 30, 2013

x(x+y)=9 ......eqn 1

y(x+y)=16 ......eqn 2

eqn 1 + eqn 2

we get:

x^2 + 2 x*y + y^2 = 9+16

(x+y)^2=25

(x + y)=sqrt(25)

x+y=5 .....eqn 3

sub eqn 3 in eqn 1

we get:

x(5)=9

x=9/5

sub eqn 3 in eqn 2

we get:

y(5)=16

y=16/5

to find: x*y

x y=(9/5) (16/5)

x*y=144/25 ......eqn 4

given that x*y=a/b .....eqn 5

comparing eqn 4 and eqn 5

we get: a=144 , b=25

(a+b)=144+25

(a+b)=169

ANS 169

Use Latex please...it's hard to understand without proper formatting

Akshat Prakash - 7 years, 7 months ago
Snehdeep Arora
Oct 28, 2013

From the first equation we get ( x + y ) = 9 x (x+y)=\frac{9}{x} putting this in second equation we get y = 16 x 9 y=\frac{16x}{9} putting this again in the second equation x = + 9 5 x=+-\frac{9}{5} and y = + 16 5 y=+-\frac{16}{5} .As x x and y y should be of the same sign to satisfy the equations x y = 144 25 xy=\frac{144}{25}

Simple substitutions can be more shorter than finding values. I hope you get my point.

Priyansh Sangule - 7 years, 7 months ago

x(x+y) = 9 y(x+y) = 16

x^2 + xy = 9 y^2 + xy = 16

adding the two eq.

(x+y)^2 = 25 x + y = 5

substituting to the first eqs.

x(5) = 9 x = 9/5

y(5) = 16 y = 16/5

Hence, xy = (16/5)(9/5) = 144/25

therefore a+b = 169

Try to answer using LATEX . It will make your solution readable .

Priyansh Sangule - 7 years, 7 months ago

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