1. Solving Systems Algebraically
Adapted from Walch Education

2. Key Concepts
Substitution is the replacement of a term of an equation by another that is known to have the same value.
When solving a quadratic-linear system, if both functions are written in function form such as “y =” or “f(x) =”, set the equations equal to each other.
5.4.2: Solving Systems Algebraically

3. Practice # 1 Solve the given system of equations algebraically.
5.4.2: Solving Systems Algebraically

4. Solution
Since both equations are equal to y, substitute by setting the equations equal to each other.
–3x + 12 = x2 – 11x + 28
Substitute –3x + 12 for y in the first equation.
5.4.2: Solving Systems Algebraically

5. Solution, continued
Solve the equation either by factoring or by using the quadratic formula.
–3x + 12 = x2 – 11x + 28
Equation
0 = x2 – 8x + 16
Set the equation equal to 0 by adding 3x to both sides, and subtracting 12 from both sides.
0 = (x – 4)2
Factor.
5.4.2: Solving Systems Algebraically

6. Solution, continued x – 4 = 0 Set each factor equal to 0 and solve.
Substitute the value of x into the second equation of the system to find the corresponding y-value.
For x = 4, y = 0. Therefore, (4, 0) is the solution.
y = –3(4) + 12
Substitute 4 for x.
y = 0
5.4.2: Solving Systems Algebraically

7. Check solution(s) by graphing
5.4.2: Solving Systems Algebraically

8. You Try! Solve the given system of equations algebraically.
5.4.2: Solving Systems Algebraically

9. Thanks For Watching!
Ms. Dambreville