Simultaneous Equations Part 2

This page expands on simultaneous equation, however you will still apply what you learned on the previous page, do not think that simultaneous equations are hard, as if you understood the previous page on simultaneous equations you will be able to understand this.
It's not necessary that both equations are the same type, for example on the previous page, each question had 2 LINEAR equations.
Lets work out the question below.
Question: Solve the simultaneous equations
  1. y2 + 2x - x2 = 25
  2. x + y = 4
Remember the aim is to solve for x or y first, we can pick what what we want to solve first and once we attain the value of either x or y, we can substitute that value into either equation to solve for the other. We can see that we can rearrange equation 2 to make either x or y the subject. Lets CHOOSE to make y the subject.x + y = 4Subtract x from both sidesx + y - x = 4 - xy = 4 - xSo now we know what y equals, we can substitute (4 - x) into equation 1.y = x - 4,soy2 + 2x - x2 = 25(Lets substitute y with (x - 4)(x – 4)2 + 2x – x2 = 25(Expand the brackets)x2 - 4x - 4x + 16 + 2x - x2 = 25Simplify-6x + 16 = 25So now we can solve to find x.-6x + 16 = 25(Subtract 16 from both sides)-6x + 16 - 16 = 25 - 16-6x = 9Divide both sides by -6-6x/-6 = 9/-6x = 9/-6 = 3/-2
Given we know have the value of x, we can substitute x with 3/-2 in either equations (1 or 2) to find the value for y. Lets choose to substitute the value in equation 1.y2 + 2(3/-2) - (3/-2)2 = 25y2 -3 - 9/4 = 25Add 9/4 to both sides y2 -3 - 9/4 + 9/4 = 25 + 9/4y2 -3 = 25 + 9/4
Add 3 to both sides
y2 -3 + 3 = 25 + 9/4 + 3y2 = 25 + 9/4 + 3y2 = 28 + 9/4y2 = 28/1 + 9/4 = (112 + 9)/4y2 = 121/4Square root both sides√y2 = √(121/4)y = ±√(121/4) (So here we have two values of y)So there are two sets of answers we get for when x is substituted with x = 3/-2, y = √(121/4)ANDx = 3/-2, y = -√(121/4)In this question, however, when we substitute x = 3/-2 and y = √(121/4) into equation 2, equation 2 is satisfied, however, when we substitute x = 3/-2 and y = -√(121/4) into equation 2, equation 2 is not satisfied. Lets show this below.
Equation 2. x + y = 4,
lets substitute x with 3/-2 and y = -√(121/4).
3/-2 + -√(121/4) = -1.5 - 5.5 = -7 ≠ 4
So x = 3/-2, y = -√(121/4) cannot be the answer as we need to find the x and y value which SIMULATENOUSLY satisfy BOTH EQUATIONS.
So our ANSWER IS
x = 3/-2, y = √(121/4)