John and Maria solved the same equation using two different meth John: Maria: 2(y + 5) = 18- 12 2y + 10 = 18 12 2(y + 5) = 18- 12 2(y+5) = 6 2 2y + 10 = 6 -10 -10 y + 5 = 3 -5 5 2y =- 4 2 2 y = -2 y = -2
John and Maria solved the same equation using two different meth John: Maria: 2(y + 5) = 18- 12 2y + 10 = 18 12 2(y + 5) = 18- 12 2(y+5) = 6 2 2y + 10 = 6 -10 -10 y + 5 = 3 -5 5 2y =- 4 2 2 y = -2 y = -2
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
Question
Verify that y=-2 is the correct solution to the equation using substitution.
Compare and contrast John and Maria's methids.Explain how they arrived at the same solution.Thank you
![John and Maria solved the same equation using two different methods.
John:
Maria:
2(y +5) = 18 - 12
2y + 10 = 18 - 12
2y + 10 = 6
-10 -10
2(y +5) = 18- 12
2fy+5) = 6
2
2
y + 5= 3
5-5
2y
=- 4
2
y= -2
y= -2
search
8.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6abd3c43-2ed8-4d20-8ecd-20e2a0fa656c%2Fc0255514-5959-4081-a5d9-ce9cb073b867%2Fivcncjq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:John and Maria solved the same equation using two different methods.
John:
Maria:
2(y +5) = 18 - 12
2y + 10 = 18 - 12
2y + 10 = 6
-10 -10
2(y +5) = 18- 12
2fy+5) = 6
2
2
y + 5= 3
5-5
2y
=- 4
2
y= -2
y= -2
search
8.
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