764 CHAPTER 11 Systems of Equations and Inequalities Solution Our plan is to eliminate x from equations (2) and (3). Multiply equation (1) by −2, and add the result to equation (2). Also, multiply equation (1) by −4, and add the result to equation (3). x y z x y z 2 8 2 3 23 − − = − + = (1) Multiply by 2. − (2) − + + =− − + = + = x y z x y z y z 2 4 2 16 2 3 23 3 7 (1) (2) Add. x y z x y z 2 8 4 5 5 53 − − = − + = (1) Multiply by 4. − (3) − + + =− − + = + = x y z x y z y z 4 8 4 32 4 5 5 53 3 9 21 (1) (3) Add. − − = + = + = ⎧ ⎨ ⎪⎪ ⎪⎪ ⎩ ⎪⎪ ⎪⎪ x y z y z y z 2 8 3 7 3 9 21 (1) (2) (3) Treat equations (2) and (3) as a system of two equations containing two variables, and eliminate the variable y by multiplying equation (2) by −3 and adding the result to equation (3). + = + = y z y z 3 7 3 9 21 (2) Multiply by 3. − (3) − − =− + = = y z y z 3 9 21 3 9 21 0 0 Add. − − = + = = ⎧ ⎨ ⎪⎪ ⎪⎪ ⎩ ⎪⎪ ⎪⎪ x y z y z 2 8 3 7 0 0 (1) (2) (3) The original system is equivalent to a system containing two equations, so the equations are dependent and the system has infinitely many solutions. If we solve equation (2) for y, we can express y in terms of z as = − + y z3 7. Substitute this expression into equation (1) to determine x in terms of z. x y z x z z x z z x z x z 2 8 2 3 7 8 6 14 8 5 22 5 22 ( ) − − = − − + − = + − − = + = = − + Equation (1) Substitute y z3 7. =− + Multiply out. Combine like terms. Solve for x. Now Work PROBLEM 47 Two distinct points in the Cartesian plane determine a unique line. Given three noncollinear points, we can find the unique quadratic function whose graph contains these three points.The next example uses a system of equations to illustrate this idea. We will write the solution to the system as = − + = − + ⎧ ⎨ ⎪⎪ ⎩ ⎪⎪ x z y z z 5 22 3 7 where can be any real number. This way of writing the solution makes it easier to find specific solutions. To find specific solutions, choose any value of z and use the equations = − + x z5 22 and = − + y z3 7 to determine x and y. For example, if = z 0, then = x 22 and = y 7, and if = z 1, then = x 17 and = y 4. Using ordered triplets, the solution is x y z x z y z z , , 5 22, 3 7, any real number ( ) { } = − + = − + Curve Fitting Find real numbers a, b, and c so that the graph of the quadratic function = + + y ax bx c 2 contains the points ( ) − − 1, 4, ( ) 1, 6 , and ( ) 3, 0 . EXAMPLE 12 Solution The three points must satisfy the equation = + + y ax bx c. 2 a b c a b c a b c a b c a b c a b c For the point 1, 4 we have: For the point 1, 6 we have: For the point 3, 0 we have: 4 1 1 6 1 1 0 3 3 4 6 0 9 3 2 2 2 − ( ) ( ) ( ) ( ) ( ) − − = − + − + = ⋅ + ⋅ + = ⋅ + ⋅ + − = − + = + + = + +
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