Solving Systems Algebraically 1 Solving Systems Algebraically WarmUp

Solving Systems Algebraically 1 Solving Systems Algebraically

Warm-Up 2 Solving Systems Algebraically

A plane flies 900 miles in 3 hours with a tail wind. It takes the same plane 5 hours to fly 900 miles with a head wind. The system expresses the distance of the plane where r is the speed of the plane in still air and w is the speed of the wind. 1. Use substitution to solve the system of equations. 2. Explain your solution(s). 3 Solving Systems Algebraically

1. Use substitution to solve the system of equations. Simplify the equations by applying the Distributive Property in both equations. 900 = (r + w)3 900 = (r – w)5 900 = 3 r + 3 w 900 = 5 r – 5 w Isolate a variable in one equation to substitute into the other equation. Let’s solve for r in the first equation. 900 = 3 r + 3 w 900 – 3 w = 3 r 300 – w = r 4 Solving Systems Algebraically

Substitute 300 – w for r in the second equation. 5 Solving Systems Algebraically

Substitute the value of w into the isolated equation to find the value of r. 300 – w = r 300 – 60 = r 240 = r 6 Solving Systems Algebraically

2. Explain your solution(s). The speed of the plane in still air is 240 miles per hour. The speed of the wind is 60 miles per hour. 7 Solving Systems Algebraically

Instruction 8 Solving Systems Algebraically

Introduction Previously, we learned how to solve quadratic-linear systems by graphing and identifying points of intersection. In this lesson, we will focus on solving a quadratic-linear system algebraically. When doing so, substitution is often the best choice. Substitution is the replacement of a variable of an equation by a number or an expression that is known to have the same value. 9 Solving Systems Algebraically

Key Concepts • When solving a quadratic-linear system, if both equations are written in function form, such as “y =” or “f(x) =, ” set the variable expressions of each equation equal to each other. • When you set the two expressions equal to each other, you are replacing y in one equation with an equivalent expression, thus using the substitution method. • If one equation is solved for y and the other is not, substitute the expression that is equal to y into the second equation. 10 Solving Systems Algebraically

Key Concepts, continued You can solve a quadratic equation by factoring or by using the quadratic formula, a formula that states that the solutions of a quadratic equation of the form ax 2 + bx + c = 0 are given by 11 Solving Systems Algebraically

Common Errors/Misconceptions • miscalculating signs • incorrectly distributing coefficients 12 Solving Systems Algebraically

Guided Practice Example 1 Solve the given system of equations algebraically. 13 Solving Systems Algebraically

Guided Practice: Example 1, continued 1. Since both equations are equal to y, substitute by setting the equations equal to each other. – 3 x +12 = x 2 − 11 x + 28 Substitute – 3 x +12 for y in the first equation. 14 Solving Systems Algebraically

Guided Practice: Example 1, continued 2. Solve the equation either by factoring or by using the quadratic formula. Rearrange the terms so that the equation is equal to 0, then solve by factoring. – 3 x +12 = x 2 − 11 x + 28 Equation 0 = x 2 − 8 x + 16 Set the equation equal to 0 by adding 3 x to both sides and subtracting 12 from both sides. 0 = (x − 4)2 Factor the trinomial Solving Systems Algebraically 15

Guided Practice: Example 1, continued x– 4=0 Take the square root of both sides. x=4 Add 4 to both sides. Substitute the value of x into the second equation of the system to find the corresponding y-value. y = – 3(4)+12 Substitute 4 for x. y=0 When x = 4, y = 0. Therefore, (4, 0) is the solution. Solving Systems Algebraically 16

Guided Practice: Example 1, continued 3. Check your solution(s) by graphing. (4, 0) 17 Solving Systems Algebraically

Guided Practice: Example 1, continued The graphs of the functions appear to intersect at (4, 0); therefore, (4, 0) checks out as the solution to this system. The solution can also be checked by substituting the coordinates (4, 0) into the given equations and solving to confirm that the results are true statements. ✔ 18 Solving Systems Algebraically

Guided Practice Example 2 Solve the given system of equations algebraically. 19 Solving Systems Algebraically

Guided Practice: Example 2, continued 1. Since both equations are equal to y, substitute by setting the equations equal to each other. x – 1 = 2 x 2 + 13 x + 15 Substitute x – 1 for y in the first equation. 20 Solving Systems Algebraically

Guided Practice: Example 2, continued 2. Solve the equation either by factoring or by using the quadratic formula. Since the equation can be factored easily, choose this method. x – 1 = 2 x 2 + 13 x + 15 Equation 0 = 2 x 2 + 12 x + 16 Set the equation equal to 0 by subtracting x from both sides and then adding 1 to both sides. 0 = 2(x + 2)(x + 4) Factor Solving Systems Algebraically 21

Guided Practice: Example 2, continued Next, set the factors equal to 0 and solve. x+2=0 x+4=0 x = – 2 x = – 4 22 Solving Systems Algebraically

Guided Practice: Example 2, continued Substitute each of the values you found for x into the second equation of the system to find the corresponding yvalue. y = (– 2) – 1 Substitute – 2 for x. y = – 3 y = (– 4) – 1 y = – 5 Substitute – 4 for x. For x = – 2, y = – 3, and for x = – 4, y = – 5. Therefore, (– 2, – 3) and (– 4, – 5) are the solutions to the system. Solving Systems Algebraically 23

Guided Practice: Example 2, continued 3. Check your solution(s) by graphing. (– 4, – 5) (– 2, – 3) 24 Solving Systems Algebraically

Guided Practice: Example 2, continued The graphs of the functions appear to intersect at ( – 2, – 3) and (– 4, – 5); therefore, (– 2, – 3) and (– 4, – 5) check out as the solutions to this system. The solutions can also be verified by substituting the coordinates into the original equations and solving to confirm that the results are true statements. ✔ 25 Solving Systems Algebraically

Guided Practice: Example 2, continued 26 Solving Systems Algebraically

Guided Practice Example 3 Solve the given system of equations algebraically. 27 Solving Systems Algebraically

Guided Practice: Example 3, continued 1. Since both equations are equal to y, substitute by setting the equations equal to each other. – 3 x – 30 = 7 x 2 − 25 x – 12 Substitute – 3 x – 30 for y in the first equation. 28 Solving Systems Algebraically

Guided Practice: Example 3, continued 2. Solve the equation either by factoring or by using the quadratic formula. The quadratic equation cannot easily be factored, so use the quadratic formula, – 3 x +12 = x 2 − 11 x + 28 Equation 0 = 7 x 2 − 22 x + 18 Set the equation equal to 0 by adding 3 x and 30 to both sides; then apply the quadratic formula 29 Solving Systems Algebraically

a = 7, b = – 22, c = 18 Use the standard form of a quadratic equation, ax 2 + bx + c = 0, to find a, b, and c. Substitute values for a, b, and c into the quadratic formula. Simplify 30 Solving Systems Algebraically

Since the value under the square root is negative, there are no real solutions to the system. Simplify to find the complex solutions. 31 Solving Systems Algebraically

Guided Practice: Example 3, continued 3. Check your solution(s) by graphing. 32 Solving Systems Algebraically

Guided Practice: Example 3, continued 3. Check your solution(s) by graphing. Since the functions do not intersect, a complex solution is feasible. ✔ 33 Solving Systems Algebraically