Questions: Solving a System of a Linear and a Quadratic Equation Algebraically Solving a system of a linear equation and a quadratic equation algebraically is similar to solving a system of linear equations. You have used substitution to solve systems of linear equations. This same strategy can be used to solve a system of a linear and a quadratic equation. Continue to consider this system of equations: Linear equation: -2x + y = 8 Quadratic equation: x^2 + y = 16 Complete the steps in the problem to solve the system algebraically. The linear equation can be written in slope-intercept form as y = 2x + 8. Substitute the expression equal to y into the quadratic equation. x^2 + (2x + 8) = 16 Subtract 16 from both sides to rewrite the equation so that it is equal to 0. x^2 + 2x - 8 = 0 Factor x^2 + 2x - 8 Now, the equation can be rewritten as (x - 2)(x + 4) = 0. Set each factor equal to 0 and solve for x. x =

Solving a System of a Linear and a Quadratic Equation Algebraically Solving a system of a linear equation and a quadratic equation algebraically is similar to solving a system of linear equations. You have used substitution to solve systems of linear equations. This same strategy can be used to solve a system of a linear and a quadratic equation. Continue to consider this system of equations: Linear equation: -2x + y = 8 Quadratic equation: x^2 + y = 16 Complete the steps in the problem to solve the system algebraically. The linear equation can be written in slope-intercept form as y = 2x + 8. Substitute the expression equal to y into the quadratic equation. x^2 + (2x + 8) = 16 Subtract 16 from both sides to rewrite the equation so that it is equal to 0. x^2 + 2x - 8 = 0 Factor x^2 + 2x - 8 Now, the equation can be rewritten as (x - 2)(x + 4) = 0. Set each factor equal to 0 and solve for x. x =
Transcript text: Solving a System of a Linear and a Quadratic Equation Algebraically Solving a system of a linear equation and a quadratic equation algebraically is similar to solving a system of linear equations. You have used substitution to solve systems of linear equations. This same strategy can be used to solve a system of a linear and a quadratic equation. Continue to consider this system of equations: Linear equation: $-2 x+y=8$ Quadratic equation: $x^{2}+y=16$ Complete the steps in the problem to solve the system algebraically. The linear equation can be written in slope-intercept form as $y=2 x+8$. Substitute the expression equal to $y$ into the quadratic equation. \[ x^{2}+\square=16 \] Subtract 16 from both sides to rewrite the equation so that it is equal to 0 . \[ x^{2}+2 x-\quad=0 \] Factor $x^{2}+2 x-8$ $\square$ Now, the equation can be rewritten as $(x-2)(x+4)=0$. Set each factor equal to 0 and solve for $x$. \[ x= \]
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Solution

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Solution Steps

Step 1: Substitute the linear equation into the quadratic equation

Given the linear equation y = 2x + 8 and the quadratic equation x² + y = 16, substitute y in the quadratic equation:

x² + (2x + 8) = 16

Step 2: Rewrite the equation to equal zero.

Subtract 16 from both sides of the equation obtained in the previous step:

x² + 2x + 8 - 16 = 0

x² + 2x - 8 = 0

Step 3: Factor the quadratic equation

Factor the quadratic equation obtained in the previous step:

(x + 4)(x - 2) = 0

Final Answer:

x = 2 or x = -4

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