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Lets rewrite our two equestion,

x²+y²=29.....i

y-x=3..........ii

Next we take our equestion ii and equet y in terms of x as follows,

y-x=3

y=x+3... Now we substitute y term in equestion i

:. x²+y²=29 where there's value of y we replace it with x+3

x²+(x+3) (x+3) =29

we rewrite the equestion in quadratic format

x²+x²+3x+3x+9=29

2x²+6x+9=29

2x²+6x+9-29=0

2x²+6x-20=0......now we have formed our quadratic equestion

To solve the equestion we may use factorisation method or completing square method. So in our case here,  I prefer to use completing square method because the our equestion doesn't have perfect squeres. Which is    [- b±(b²_4ab)¹/²]/2a

Next we substitute the coefficients of x in our equestion above,

where a=2, b=6, and c=-20

[-(6)±(6²-{4×2×-20})¹/²]/2×2

{-6±(36+160)¹/²}/4

{-6±(196)¹/²}/4 where the root of 196=14

(-6±14)/4

x=(-6+14)/4=2

x=(-6-14)/4=-5

Now we have the possible values of x 2&-5, then we substitute these values on our y equestion

y=x+3

y=2+3

y=5

y=-5+3

y=-2

Therefore our values for both x and y will be

x=2 or - 5

y=5 or - 2 This is because our simultaneous equestion had a quadratic equestion.

You can solve it by, first, isolate y in the second equation:

• y-x=3
• y=3-x [Equation 1]

Second, get the square of the resulting [Equation 1]:

• y^2 = 9 - 6x + x^2 [Equation 2]

Substitute, [Equation 2] to the first equation:

• x^2 + y^2 = 29
• x^2 + 9 - 6x + x^2 = 29 [Equation 3]

Simplify [Equation 3] into a quadratic equation:

• x^2 + 9 - 6x + x^2 = 29
• 2x^2 - 6x + 9 - 29 = 0
• 2x^2 - 6x - 20 = 0 [Equation 4]

Substitute the following values into the quadratic equation:

• 2x^2 - 6x - 20 = 0 [Equation 4]
• a = 2
• b = -6
• c = -20

• x = 5 and/or
•  x=-2
• (NOTE that there are two answers since it is a quadratic equation).

To solve for y, simply use the second equation above:

• Use x = 5: then, y - 5 = 3
• Use x= -2: then, y - (-2) = 3

• x = 5 and  y = 8
• x = -2 and y = 1

(NOTE that there are two sets of answers because the original equations are in quadratic form)

You can solve it by, first, isolate y in the second equation:

• y-x=3
• y=3-x [Equation 1]

Second, get the square of the resulting [Equation 1]:

• y^2 = 9 - 6x + x^2 [Equation 2]

Substitute, [Equation 2] to the first equation:

• x^2 + y^2 = 29
• x^2 + 9 - 6x + x^2 = 29 [Equation 3]

Simplify [Equation 3] into a quadratic equation:

• x^2 + 9 - 6x + x^2 = 29
• 2x^2 - 6x + 9 - 29 = 0
• 2x^2 - 6x - 20 = 0 [Equation 4]

Substitute the following values into the quadratic equation:

• 2x^2 - 6x - 20 = 0 [Equation 4]
• a = 2
• b = -6
• c = -20

• x = 5 and/or
•  x=-2
• (NOTE that there are two answers since it is a quadratic equation).

To solve for y, simply use the second equation above:

• Use x = 5: then, y - 5 = 3
• Use x= -2: then, y - (-2) = 3

• x = 5 and  y = 8
• x = -2 and y = 1

(NOTE that there are two sets of answers because the original equations are in quadratic form)

Then, substitute the squared equa

You can solve it by, first, isolate y in the second equation:

• y-x=3
• y=3-x [Equation 1]

Second, get the square of the resulting [Equation 1]:

• y^2 = 9 - 6x + x^2 [Equation 2]

Substitute, [Equation 2] to the first equation:

• x^2 + y^2 = 29
• x^2 + 9 - 6x + x^2 = 29 [Equation 3]

Simplify [Equation 3] into a quadratic equation:

• x^2 + 9 - 6x + x^2 = 29
• 2x^2 - 6x + 9 - 29 = 0
• 2x^2 - 6x - 20 = 0 [Equation 4]

Substitute the following values into the quadratic equation:

• 2x^2 - 6x - 20 = 0 [Equation 4]
• a = 2
• b = -6
• c = -20

• x = 5 and/or
•  x=-2
• (NOTE that there are two answers since it is a quadratic equation).

To solve for y, simply use the second equation above:

• Use x = 5: then, y - 5 = 3
• Use x= -2: then, y - (-2) = 3

• x = 5 and  y = 8
• x = -2 and y = 1

(NOTE that there are two sets of answers because the original equations are in quadratic form)

Then, substitute the squared equa

Hi

Hope it helps

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

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