Splash Screen. Lesson Menu Five-Minute Check (over Lesson 10–6) Then/Now Example 1:...
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Transcript of Splash Screen. Lesson Menu Five-Minute Check (over Lesson 10–6) Then/Now Example 1:...
Five-Minute Check (over Lesson 10–6)
Then/Now
Example 1: Linear-Quadratic System
Example 2: Quadratic-Quadratic System
Example 3: Quadratic Inequalities
Example 4: Quadratics with Absolute Value
You solved systems of linear equations. (Lessons 3–1 and 3–2)
• Solve systems of linear and nonlinear equations algebraically and graphically.
• Solve systems of linear and nonlinear inequalities graphically.
Linear-Quadratic System
Solve the system of equations.4x2 – 16y2 = 25 (1)2y + x = 2 (2)
Step 1 Solve the linear equation for y.
2y + x = 2 Equation (2)
Simplify.
Solve for y.
Linear-Quadratic System
Step 2 Substitute into the quadratic equation and solve for x.
Quadratic equation
Substitute – x + 1 for y.__1
2
Simplify.
4x2 – 4x2 + 16x – 16 = 25 Distribute.
Linear-Quadratic System
16x = 41 Add 16 to each side.
Divide each side by 16.
Step 3 Substitute x into the linear equation and solve for y.
Linear equation
Substitute for x.
Linear-Quadratic System
Answer: The solution is .
Simplify.
A. A
B. B
C. C
D. D
What is the solution to the system of equations?x2 – y2 = 42y + x = 2
A.
B.
C.
D.
Quadratic-Quadratic System
Solve the system of equations.x2 + y2 = 16 (1)4x2 + y2 = 23 (2)
x2 + y2 = 16Equation (1)
(–) 4x2 + y2= 23Equation (2)
–3x2 = –7Subtract.
Divide each side –3.
Take the square root of each side.
Quadratic-Quadratic System
Multiply by a form of 1.
Simplify.
Substitute into one of the original
equations and solve for y.
Equation (1)
Substitute for x.
Quadratic-Quadratic System
Simplify.
Subtract
from each side.
Take the square root of each side.
Quadratic-Quadratic System
Answer: The solutions are
A. A
B. B
C. C
D. D
Solve the system of equations.x2 + y2 = 36 (1)x2 + 3y2 = 42 (2)
A.
B.
C.
D.
Quadratic Inequalities
Solve the system of inequalities by graphing.
y > x2 + 1
x2 + y2 ≤ 9
The graph of y > x2 + 1 is the parabola y = x2 + 1 and the region inside and above it. The region is shaded blue.
The graph of x2 + y2 ≤ 9 is the interior of the circle x2 + y2 = 9. This region is shaded yellow.
Quadratic Inequalities
The intersection of these regions, shaded green, represents the solution of the system of inequalities.
Answer:
Check (0, 2) is in the shaded area. Use this point to check your solution.
y > x2 + 1 x2 + y2 ≤ 9
2 > 1 4 ≤ 9
2 > 02 + 1 02 + 22 ≤ 9? ?
A. A
B. B
C. C
D. D
Solve the system of inequalities by graphing.y < –x2 + 1x2 + y2 ≤ 4
A. B.
C. D.
Quadratics with Absolute Value
Solve the system of inequalities by graphing.
x2 + y2 > 16
y > │x + 4│
Graph the boundary equations. Then shade appropriately.
The intersection of the graphs, shaded green, represents the solution to the system.
x2 + y2 > 16 y > │x + 4│
Quadratics with Absolute Value
Check (–4, 4) is in the shaded area. Use the points to
check your solution.
32 > 16 4 > 0
? ?(–4)2 + 42 > 16 4 > │–4 + 4│
A. A
B. B
C. C
D. D
Solve the system of inequalities by graphing.x2 + y2 > 9y < │x + 1│+ 1
A. B.
C. D.
