{"id":9202,"date":"2020-12-16T12:43:44","date_gmt":"2020-12-16T07:13:44","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=9202"},"modified":"2020-12-16T15:25:13","modified_gmt":"2020-12-16T09:55:13","slug":"solving-linear-quadratic-systems-algebraically","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/solving-linear-quadratic-systems-algebraically\/","title":{"rendered":"Solving Linear Quadratic Systems Algebraically"},"content":{"rendered":"

Solving Linear Quadratic Systems Algebraically<\/strong><\/span><\/h2>\n

A linear quadratic system is a system containing one linear equation and one quadratic equation
\nwhich may be one straight line and one parabola,
\nor one straight line and one circle.
\nAlgebraic Solutions
\nstraight line: y = mx + b
\nparabola: y = ax2<\/sup> + bx + c; a \u2260 0
\ncircle: (x – h)2<\/sup> + (y – k)2<\/sup> = r2<\/sup> ; center (h.k), radius r<\/p>\n

Let’s look at how to solve a linear quadratic system of equations algebraically.<\/span><\/strong><\/p>\n

Example 1:
\nSolve this linear-quadratic system of equations algebraically and check your solution:
\ny = x2<\/sup>– 6x + 3 (parabola)
\ny = -2x + 3 (straight line)<\/strong>
\nSolution:<\/strong><\/p>\n

    \n
  1. Solve for one of the variables in the linear equation.
    \nNote: In this example, this process is already done for us, since y = -2x + 3.
    \ny = -2x + 3<\/li>\n
  2. Substitute this value into the quadratic equation, and solve the resulting equation.
    \n\u2022 Substitute -2x + 3 for y in the quadratic equation.
    \n\u2022 Subtract 3 from both sides; then add 2x to both sides.
    \n\u2022 Factor.
    \n\u2022 Set each factor equal to zero and solve.
    \nYou now have TWO values for x. This tells you that there may be two possible solutions.
    \nTWO SOLUTIONS.
    \n\"Solving<\/li>\n
  3. Find the corresponding values for y. Substitute each value into the linear equation in place of x.
    \nYes, you could substitute in the quadratic equation, but substituting into the linear equation will be easier.
    \n\"Solving<\/li>\n
  4. Check: Be sure to check BOTH solutions in both equations.
    \n\"Solving<\/li>\n
  5. State the final solutions. The solutions may be stated as the set {(0, 3), (4, -5)}<\/li>\n<\/ol>\n

    Example 2:
    \nSolve this linear-quadratic system of equations algebraically and check your solution:
    \ny = x2<\/sup>– 6x + 3 (parabola)
    \n2x – y = 13 (straight line)<\/strong>
    \nSolution:<\/strong><\/p>\n

      \n
    1. Solve for one of the variables in the linear equation.
      \n2x – y = 13
      \ny = 2x – 13<\/li>\n
    2. Substitute this value into the quadratic equation, and solve the resulting equation.
      \n\u2022 Substitute 2x – 13 for y in the quadratic equation.
      \n\u2022 Add 13 to both sides; then subtract 2x from both sides.
      \n\u2022 Factor.
      \n\u2022 Set each factor equal to zero and solve.
      \nYou now have ONE value for x. This tells you that there may be only one solution.
      \n\"Solving<\/li>\n
    3. Find the corresponding value for y. Substitute the value into the linear equation in place of x.
      \nYes, you could substitute in the quadratic equation, but substituting into the linear equation will be easier.
      \n\"Solving<\/li>\n
    4. Check: Be sure to check the solution in both equations.
      \n\"Solving<\/li>\n
    5. State the final solution. The solution may be stated as (4, -5) or {(4, -5)}<\/li>\n<\/ol>\n

      Example 3:<\/strong>
      \nSolve this linear-quadratic system of equations algebraically and check your solution:<\/strong>
      \nx2<\/sup>+ y2<\/sup> = 9 (circle)<\/strong>
      \nx – y = 3 (straight line)
      \nSolution:<\/strong><\/p>\n

        \n
      1. Solve for one of the variables in the linear equation.
        \ny\u00a0=\u00a0<\/strong>x\u00a0– 3<\/li>\n
      2. Substitute this value into the quadratic equation, and solve the resulting equation.
        \n\u2022 Substitute x – 3 for y in the quadratic equation.
        \n\u2022 Expand (x – 3)2<\/sup>
        \n\u2022 Combine terms.
        \n\u2022 Factor.
        \n\u2022 Set each factor equal to zero and solve.
        \nYou now have TWO values for x. This tells you that there may be two possible solutions.
        \n\"Solving<\/li>\n
      3. Find the corresponding values for y. Substitute each value into the linear equation in place of x.
        \nYes, you could substitute in the quadratic equation, but substituting into the linear equation will be easier.
        \n\"Solving<\/li>\n
      4. Check: Be sure to check BOTH solutions in both equations.
        \n\"Solving<\/li>\n
      5. State the final solutions. The solutions may be stated as the set {(0, -3), (3, 0)}<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

        Solving Linear Quadratic Systems Algebraically A linear quadratic system is a system containing one linear equation and one quadratic equation which may be one straight line and one parabola, or one straight line and one circle. Algebraic Solutions straight line: y = mx + b parabola: y = ax2 + bx + c; a \u2260 … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[3185,3361,3360],"yoast_head":"\nSolving Linear Quadratic Systems Algebraically - CBSE Library<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/cbselibrary.com\/solving-linear-quadratic-systems-algebraically\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Solving Linear Quadratic Systems Algebraically\" \/>\n<meta property=\"og:description\" content=\"Solving Linear Quadratic Systems Algebraically A linear quadratic system is a system containing one linear equation and one quadratic equation which may be one straight line and one parabola, or one straight line and one circle. Algebraic Solutions straight line: y = mx + b parabola: y = ax2 + bx + c; a \u2260 ... 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