{"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":"
A linear quadratic system is a system containing one linear equation and one quadratic equation Let’s look at how to solve a linear quadratic system of equations algebraically.<\/span><\/strong><\/p>\n Example 1: Example 2: Example 3:<\/strong> 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":"\n
\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
\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
\nNote: In this example, this process is already done for us, since y = -2x + 3.
\ny = -2x + 3<\/li>\n
\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<\/li>\n
\nYes, you could substitute in the quadratic equation, but substituting into the linear equation will be easier.
\n<\/li>\n
\n<\/li>\n
\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
\n2x – y = 13
\ny = 2x – 13<\/li>\n
\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<\/li>\n
\nYes, you could substitute in the quadratic equation, but substituting into the linear equation will be easier.
\n<\/li>\n
\n<\/li>\n
\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
\ny\u00a0=\u00a0<\/strong>x\u00a0– 3<\/li>\n
\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<\/li>\n
\nYes, you could substitute in the quadratic equation, but substituting into the linear equation will be easier.
\n<\/li>\n
\n<\/li>\n