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**10.7 Solving Quadratic Systems**

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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**

x2 + y2 = & y = x + 1 We will use….. Substitution. x2 + (x + 1)2 = 13 x2 + (x2 + 2x + 1) = 13 2x2 + 2x – 12 = 0 2(x2 + 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:**

x2 + y2 = 5 & y = -x + 3 (1,2) & (2,1)

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**Solve by substitution: x2 + 4y2 – 4 = 0 -2y2 + x + 2 = 0**

The second equation has no x2 term so solve for x → x = 2y2 – 2 and substitute it into the first equation. (2y2 – 2)2 + 4y2 - 4 = 0 4y4 – 8y y2 – 4 = 0 4y4 – 4y2 = 0 4y2 (y2 – 1) = 0 4y2 (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 x2 + y2 – 16x + 39 = 0 x2 – y2 – 9 = 0**

If you add these two equations together, the y’s will cancel x2 – y = 0

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x2 + y2 – 16x + 39 = 0 x2 – y = 0 2x – 16x + 30 = 0 2(x2 – 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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Algebra Review. Systems of Equations Review: Substitution Linear Combination 2 Methods to Solve:

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