Solving Systems Algebraically Adapted from Walch Education

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

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

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

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

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

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

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

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