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Hello (matt)

Since there are squared terms you can get two answers for both x and y, don't let this throw you!

The first step is to make x or y the subject of the simpler equation. Then we can substitute into the more complicated one. I'll make y the subject of the simple equation...

y-x=3

y=3+x

 

Now substitute this into the other equation:

X^2 + y^2 = 29

 

x^2 + (3+x)^2 = 29

x^2 + (3+x)(3+x) = 29

x^2 + 9 + 6x +x^2 = 29

 

We now want to solve this equation for x:

x^2 + 9 + 6x +x^2 = 29

2x^2 +6x = 20

2x^2 +6x -20 = 0

(2x-4)(x+5)=0

 

x= 2 or x =-5

 

We can now use these solutions for x to find the solutions for y by substituting them into the simpler equation, I’ll start with x =2

 

y-x=3

y-2=3

y=5

 

so y=5 when x =2

 

or

 

y-x=3

y-(-5)=3

y+5=3

y=-2

 

y=-2 when x =-5

 answers are in bold.

From second equation y-x  =  3  <=>   y = 3+x

substitute y= 3+x into equation (1)

=> x^2 + (3+x)^2  = 29

=>  x= -5    and x=2

if x=-5   then y = 3- 5   = -2

if  x= 2  then y = 3-2 = x

let x2+y2=29-----(1)

y-x=3---(2)

in eq.2 squaring both sides

(y-x)2=(3)2   <(a-b)2= a2+b2-2ab

y2+x2-2xy= 9-----(3)

substituting value of x2+y2 in equation(3)

29-2xy= 9

29-9= 2xy

20= 2xy

10= xy

x= 10/y------(4)

substituting value of x in equation (2)

y-10/y=3

y2-10= 3y

y2-3y-10=0

splitting the equation:

y2-5y+2y-10=0

y(y-5) + 2(y-5)=0

(y+2)(y-5)=0

y= -2,5

x= -2-3= -5

x= 5-3= 2

x= -5,2

Hi Matt!

You can use substitution. Re-write the second equation like this:

y = 3 + x

Now you can replace y with 3 + x in the first equation:

x^2 + (3 + x)^2 = 29

Expand this out and simplify:

x^2 + 9 + 6x + x^2 = 29

2x^2 + 6x = 20

x^2 + 3x - 10 = 0

Solve this (you can factorise, complete the square, or use the quadratic formula):

You will find x = -5 or x = 2.

Now we can use this to find y. Using the second equation is easier:

y = 3 + (-5) or y = 3 + 2

So y = -2 or y = 5.

Final answer is x = -5, y = -2 or x = 2, y = 5.

Hope that helps!

Christophe

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