Try to figure out how to solve this equation: x2 + y2 = 29 y - x = 3

## How to Find the Answer

Lets begin by rewriting our two equations, x²+y²=29…..iy-x=3……….ii Next we take our equation ii and equate y in terms of x as follows, y-x=3y=x+3… Now we substitute y term in equation i:. x²+y²=29 where there’s the value of y we replace it with x+3 x²+(x+3) (x+3) =29 we rewrite the equation in quadratic formatx²+x²+3x+3x+9=292x²+6x+9=29 2x²+6x+9-29=02x²+6x-20=0……now we have formed our quadratic equation. To solve the equation we may use the factorization method or completing square method. So in the case here,  I prefer to use the completing square method because our equation doesn’t have perfect squares. Which is [- b±(b²_4ab)¹/²]/2a Next, we substitute the coefficients of x in our equation 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)/4x=(-6+14)/4=2x=(-6-14)/4=-5 Now we have the possible values of x 2&-5, then we substitute these values on our y equation y=x+3y=2+3y=5y=-5+3y=-2 Therefore our values for both x and y will be x=2 or – 5y=5 or – 2 This is because our simultaneous equation had a quadratic equation.

Hi Matt!You can use substitution. Re-write the second equation like this:y = 3 + xNow you can replace y with 3 + x in the first equation:x^2 + (3 + x)^2 = 29Expand this out and simplify:x^2 + 9 + 6x + x^2 = 292x^2 + 6x = 20x^2 + 3x - 10 = 0Solve 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 + 2So y = -2 or y = 5.Final answer is x = -5, y = -2 or x = 2, y = 5.Hope that helps! Christophe
Christophe J.
14 July 2017
let x2+y2=29-----(1)y-x=3---(2)in eq.2 squaring both sides(y-x)2=(3)2   <(a-b)2= a2+b2-2aby2+x2-2xy= 9-----(3)substituting value of x2+y2 in equation(3)29-2xy= 929-9= 2xy20= 2xy10= xyx= 10/y------(4)substituting value of x in equation (2)y-10/y=3y2-10= 3yy2-3y-10=0splitting the equation:y2-5y+2y-10=0y(y-5) + 2(y-5)=0(y+2)(y-5)=0y= -2,5x= -2-3= -5x= 5-3= 2x= -5,2
parulrana1992
15 July 2017
From second equation y-x  =  3  <=>   y = 3+xsubstitute y= 3+x into equation (1)=> x^2 + (3+x)^2  = 29=>  x= -5    and x=2if x=-5   then y = 3- 5   = -2if  x= 2  then y = 3-2 = x
ms181920
15 July 2017
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Sanziana F.
16 July 2017
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=3y=3+x Now substitute this into the other equation:X^2 + y^2 = 29 x^2 + (3+x)^2 = 29x^2 + (3+x)(3+x) = 29x^2 + 9 + 6x +x^2 = 29 We now want to solve this equation for x:x^2 + 9 + 6x +x^2 = 292x^2 +6x = 202x^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=3y-2=3y=5 so y=5 when x =2  or y-x=3y-(-5)=3y+5=3y=-2 y=-2 when x =-5 answers are in bold.
Victor A.
17 July 2017
Sheik A.
27 July 2017
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Sheik A.
27 July 2017
Lizette A.
29 July 2017