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Lesson 2 Menu : Lesson 2 Menu Five-Minute Check (over Lesson 5-1)
Main Ideas and Vocabulary
California Standards
Example 1: Solve Using Substitution
Example 2: Infinitely Many or No Solutions
Example 3: Write and Solve a System of Equations
Lesson 2 MI/Vocab : Lesson 2 MI/Vocab substitution Solve systems of equations algebraically by using substitution. Solve real-world problems involving systems of equations.
Lesson 2 CA : Lesson 2 CA Standard 9.0 Students solve a system of two linear equations in two variables algebraically and are able to interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets. (Key, CAHSEE)
Lesson 2 Ex1 : Lesson 2 Ex1 Solve Using Substitutions A. Use substitution to solve the system of equations.x = 4y4x – y = 75 Since x = 4y, substitute 4y for x in the second equation.
4x – y = 75 Second equation
4(4y) – y = 75 x = 4y
16y – y = 75 Simplify.
15y = 75 Combine like terms.
Lesson 2 Ex1 : Lesson 2 Ex1 Solve Using Substitutions Divide each side by 15. y = 5 Simplify.
Use x = 4y to find the value of x.
x = 4y First equation
x = 4(5) y = 5
x = 20 Simplify. Answer: The solution is (20, 5).
Lesson 2 Ex1 : Lesson 2 Ex1 Solve Using Substitutions B. Use substitution to solve the system of equations.4x + y = 12–2x – 3y = 14 Solve the first equation for y since the coefficient of y is 1.
4x + y = 12 First equation
4x + y – 4x = 12 – 4x Subtract 4x from each side.
y = 12 – 4x Simplify.
Lesson 2 Ex1 : Lesson 2 Ex1 Solve Using Substitutions Find the value of x by substituting 12 – 4x for y in the second equation. –2x – 3y = 14 Second equation
–2x – 3(12 – 4x) = 14 y = 12 – 4x
–2x – 36 + 12x = 14 Distributive Property
10x – 36 = 14 Combine like terms.
10x – 36 + 36 = 14 + 36 Add 36 to each side.
10x = 50 Simplify. x = 5 Simplify. Divide each side by 10.
Lesson 2 Ex1 : Lesson 2 Ex1 Solve Using Substitutions Substitute 5 for x in either equation to find the value of y. 4x + y = 12 First equation
4(5) + y = 12 x = 5
20 +y = 12 Simplify.
y = –8 Subtract 20 from each side. Answer: The solution is (5, –8). The graph verifies the solution.
Slide 10 : A
B
C
D A. Use substitution to solve the system of equations.y = 2x3x + 4y = 11
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B
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D B. Use substitution to solve the system of equations.x + 2y = 15x – 4y = –23
Lesson 2 Ex2 : Lesson 2 Ex2 Infinitely Many or No Solutions Use substitution to solve the system of equations.2x + 2y = 8x + y = –2 Solve the second equation for y.
x + y = –2 Second equation
x + y – x = –2 – x Subtract x from each side.
y = –2 – x Simplify.
Substitute –2 – x for y in the first equation.
2x + 2y = 8 First equation
2x + 2(–2 – x) = 8 y = –x – 2
Lesson 2 Ex2 : Lesson 2 Ex2 Infinitely Many or No Solutions 2x – 4 – 2x = 8 Distributive Property –4 = 8 Simplify. Answer: no solution The statement –4 = 8 is false. This means there are no solutions of the system of equations.
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B
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D A. one; (0, 0)
B. no solution
C. infinitely many solutions
D. cannot be determined Use substitution to solve the system of equations.3x – 2y = 3–6x + 4y = –6
Lesson 2 Ex3 : Lesson 2 Ex3 Write and Solve a System of Equations
Lesson 2 Ex3 : Lesson 2 Ex3 Write and Solve a System of Equations Let x = the number of ounces of 18-karat gold andy = the number of ounces of 12-karat gold. Use the table to organize the information.
Lesson 2 Ex3 : Lesson 2 Ex3 Write and Solve a System of Equations x + y = 4 First equation x + y – y = 4 – y Subtract y from each side.
x = 4 – y Simplify. Second equation x = 4 – y Distributive Property
Lesson 2 Ex3 : Lesson 2 Ex3 Write and Solve a System of Equations Combine like terms Subtract 3 from each side. Simplify. Multiply each side by –4. Simplify.
Lesson 2 Ex3 : Lesson 2 Ex3 Write and Solve a System of Equations x + y = 4 First equation Simplify.
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B
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D A. 0 mL of 10% solution, 10 mL of 40% solution
B. 6 mL of 10% solution, 4 mL of 40% solution
C. 5 mL of 10% solution, 5 mL of 40% solution
D. 3 mL of 10% solution, 7 mL of 40% solution CHEMISTRY Mikhail needs 10 milliliters of 25% HCl (hydrochloric acid) solution for a chemistry experiment. There is a bottle of 10% HCl solution and a bottle of 40% HCl solution in the lab. How much of each solution should he use to obtain the required amount of 25% HCl solution?
End of Lesson 2 : End of Lesson 2