Solving Quadratic Simultaneous Equations

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.  

Answers
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
HiPlease see the file attach for the answer to your question Hope it helps 
Sheik A.
27 July 2017
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Sheik A.
27 July 2017
You can solve it by, first, isolate y in the second equation:y-x=3y=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 = 29x^2 + 9 - 6x + x^2 = 29 [Equation 3]Simplify [Equation 3] into a quadratic equation:x^2 + 9 - 6x + x^2 = 292x^2 - 6x + 9 - 29 = 02x^2 - 6x - 20 = 0 [Equation 4]Use the quadratic equation formula (refer to http://dragonometry.net/blog/wp-content/uploads/2013/09/quadraticformula.jpg) to solve [Equation 4]Substitute the following values into the quadratic equation:2x^2 - 6x - 20 = 0 [Equation 4]a = 2b = -6c = -20Using the quadratic formula, the answer is: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 = 3Answer: y = 8Use x= -2: then, y - (-2) = 3Answer: y = 1Therefore, the answers are:x = 5 and  y = 8x = -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
Lizette A.
29 July 2017
You can solve it by, first, isolate y in the second equation:y-x=3y=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 = 29x^2 + 9 - 6x + x^2 = 29 [Equation 3]Simplify [Equation 3] into a quadratic equation:x^2 + 9 - 6x + x^2 = 292x^2 - 6x + 9 - 29 = 02x^2 - 6x - 20 = 0 [Equation 4]Use the quadratic equation formula (refer to http://dragonometry.net/blog/wp-content/uploads/2013/09/quadraticformula.jpg) to solve [Equation 4]Substitute the following values into the quadratic equation:2x^2 - 6x - 20 = 0 [Equation 4]a = 2b = -6c = -20Using the quadratic formula, the answer is: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 = 3Answer: y = 8Use x= -2: then, y - (-2) = 3Answer: y = 1Therefore, the answers are:x = 5 and  y = 8x = -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
Lizette A.
29 July 2017
You can solve it by, first, isolate y in the second equation:y-x=3y=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 = 29x^2 + 9 - 6x + x^2 = 29 [Equation 3]Simplify [Equation 3] into a quadratic equation:x^2 + 9 - 6x + x^2 = 292x^2 - 6x + 9 - 29 = 02x^2 - 6x - 20 = 0 [Equation 4]Use the quadratic equation formula (refer to http://dragonometry.net/blog/wp-content/uploads/2013/09/quadraticformula.jpg) to solve [Equation 4]Substitute the following values into the quadratic equation:2x^2 - 6x - 20 = 0 [Equation 4]a = 2b = -6c = -20Using the quadratic formula, the answer is: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 = 3Answer: y = 8Use x= -2: then, y - (-2) = 3Answer: y = 1Therefore, the answers are:x = 5 and  y = 8x = -2 and y = 1(NOTE that there are two sets of answers because the original equations are in quadratic form)
Lizette A.
29 July 2017
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G B.
13 August 2017
Lets rewrite our two equestion, x²+y²=29.....iy-x=3..........iiNext we take our equestion ii and equet y in terms of x as follows, y-x=3y=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 formatx²+x²+3x+3x+9=292x²+6x+9=29 2x²+6x+9-29=02x²+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]/2aNext 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)/4x=(-6+14)/4=2x=(-6-14)/4=-5Now we have the possible values of x 2&-5, then we substitute these values on our y equestion y=x+3y=2+3y=5y=-5+3y=-2Therefore our values for both x and y will be x=2 or - 5y=5 or - 2 This is because our simultaneous equestion had a quadratic equestion. 
dykes
17 August 2017
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