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10.7 Solving Quadratic Systems p. 632

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We’ve already studied two techniques for solving systems of linear equations. You will use these same techniques to solve quadratic systems. These techniques are ??? Substitution Linear combination

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Find the points of intersection x 2 + y 2 = 13 & y = x + 1 We will use….. Substitution. x 2 + (x + 1) 2 = 13 x 2 + (x 2 + 2x + 1) = 13 2x 2 + 2x – 12 = 0 2(x 2 + x – 6) = 0 2(x + 3)(x – 2) = 0 x = -3 & x = 2 Now plug these values into the Equation to get y!! (-3,-2) and (2,3) are the points where the two graphs intersect. Check it on your calculator!

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Your turn! Find the points of intersection of: x 2 + y 2 = 5 & y = -x + 3 (1,2) & (2,1)

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Solve by substitution: x 2 + 4y 2 – 4 = 0 -2y 2 + x + 2 = 0 The second equation has no x 2 term so solve for x → x = 2y 2 – 2 and substitute it into the first equation. (2y 2 – 2) 2 + 4y 2 - 4 = 0 4y 4 – 8y 2 + 4 + 4y 2 – 4 = 0 4y 4 – 4y 2 = 0 4y 2 (y 2 – 1) = 0 4y 2 (y-1)(y+1) = 0 y = 0, y = 1, y = -1 Now plug these x values into The revised equation Which gives you : (-2,0) (0,1) (0,-1)

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Linear combination x 2 + y 2 – 16x + 39 = 0 x 2 – y 2 – 9 = 0 If you add these two equations together, the y’s will cancel x 2 + y 2 – 16x + 39 = 0 x 2 – y 2 - 9 = 0

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x 2 + y 2 – 16x + 39 = 0 x 2 – y 2 - 9 = 0 2x 2 – 16x + 30 = 0 2(x 2 – 8x + 15) = 0 2 (x-3) ( x-5) = 0 x = 3 or x = 5 Plugging these into one of the ORIGINAL equations to get: (3,0) (5,4) ( 5,-4)

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