Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015

1st Edition

ISBN: 9781680331141

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Houghton Mifflin Harcourt

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Concept explainers

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.

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Question

Chapter 1, Problem 4CA

(a)

To determine

To write: An equation that represents the no of cans bought by the person.

(a)

Expert Solution

Answer to Problem 4CA

The equation for the cans is 132=24x+28(5−x) , the number of white cans is 2 and the number of blue cans is 3.

Explanation of Solution

Given:

The cost of white paint of one can is $24 for dining room and the cost of blue paint of one can is $28 for the living room. The total money spend on 5 cans is $132 .

The requirement is to find an equation by the given numbers and symbols.

Consider x as the number of white cans and 5−x as the number of blue cans.

Write an equation for the total cost of 5 cans by the provided information.

132=24x+28(5−x)

Now solve the equation for x .

132=24x+28(5−x)132=24x+140−28x132−140=−4x−8=−4x

Further simlify the above equation

8=4xx=84x=2

Therefore, the number of white cans is 2 and the number of blue cans is 3.

Conclusion:

Thus, the equation for the cans is 132=24x+28(5−x) , the number of white cans is 2 and the number of blue cans is 3.

(b)

To determine

To find: The saved money if colors are switched for the dinning rom and living room.

(b)

Expert Solution

Answer to Problem 4CA

The money saved by the person is $4 .

Explanation of Solution

Calculation:

The colors are switched that means the cans of white paint is 3 and the cans of blue paint is 2.

There is need to find the cost of 3 white cans and 2 blue cans.

Substitute x=3 in the equation 132=24x+28(5−x) to find the cost of 3 white cans and 2 blue cans.

C=24(3)+28(5−3)=24(3)+28(2)=72+56=128

The money spent by the person is $132 and the money have to pay is $128 if colors are switched.

Find the amount of saved money.

$132−$128=$4

Therefore, the money saved by the person is $4 .

Conclusion:

Thus, the money saved by the person is $4 .

Chapter 1, Problem 3CA

Chapter 1, Problem 5CA

Chapter 1 Solutions

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015

Ch. 1.1 - Prob. 1E Ch. 1.1 - Prob. 2E Ch. 1.1 - Prob. 3E Ch. 1.1 - Prob. 4E Ch. 1.1 - Prob. 5E Ch. 1.1 - Prob. 6E Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E

Ch. 1.1 - Prob. 11E Ch. 1.1 - Prob. 12E Ch. 1.1 - Prob. 13E Ch. 1.1 - Prob. 14E Ch. 1.1 - Prob. 15E Ch. 1.1 - Prob. 16E Ch. 1.1 - Prob. 17E Ch. 1.1 - Prob. 18E Ch. 1.1 - Prob. 19E Ch. 1.1 - Prob. 20E Ch. 1.1 - Prob. 21E Ch. 1.1 - Prob. 22E Ch. 1.1 - Prob. 23E Ch. 1.1 - Prob. 24E Ch. 1.1 - Prob. 25E Ch. 1.1 - Prob. 26E Ch. 1.1 - Prob. 27E Ch. 1.1 - Prob. 28E Ch. 1.1 - Prob. 29E Ch. 1.1 - Prob. 30E Ch. 1.1 - Prob. 31E Ch. 1.1 - Prob. 32E Ch. 1.1 - Prob. 33E Ch. 1.1 - Prob. 34E Ch. 1.1 - Prob. 35E Ch. 1.1 - Prob. 36E Ch. 1.1 - Prob. 37E Ch. 1.1 - Prob. 38E Ch. 1.1 - Prob. 39E Ch. 1.1 - Prob. 40E Ch. 1.1 - Prob. 41E Ch. 1.1 - Prob. 42E Ch. 1.1 - Prob. 43E Ch. 1.1 - Prob. 44E Ch. 1.1 - Prob. 45E Ch. 1.1 - Prob. 46E Ch. 1.1 - Prob. 47E Ch. 1.1 - Prob. 48E Ch. 1.1 - Prob. 49E Ch. 1.1 - Prob. 50E Ch. 1.1 - Prob. 51E Ch. 1.1 - Prob. 52E Ch. 1.1 - Prob. 53E Ch. 1.1 - Prob. 54E Ch. 1.1 - Prob. 55E Ch. 1.1 - Prob. 56E Ch. 1.1 - Prob. 57E Ch. 1.1 - Prob. 58E Ch. 1.1 - Prob. 59E Ch. 1.1 - Prob. 60E Ch. 1.1 - 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Prob. 33E Ch. 1.3 - Prob. 34E Ch. 1.3 - Prob. 35E Ch. 1.3 - Prob. 36E Ch. 1.3 - Prob. 37E Ch. 1.3 - Prob. 38E Ch. 1.3 - Prob. 39E Ch. 1.3 - Prob. 40E Ch. 1.3 - Prob. 41E Ch. 1.3 - Prob. 42E Ch. 1.3 - Prob. 43E Ch. 1.3 - Prob. 44E Ch. 1.3 - Prob. 1Q Ch. 1.3 - Prob. 2Q Ch. 1.3 - Prob. 3Q Ch. 1.3 - Prob. 4Q Ch. 1.3 - Prob. 5Q Ch. 1.3 - Prob. 6Q Ch. 1.3 - Prob. 7Q Ch. 1.3 - Prob. 8Q Ch. 1.3 - Prob. 9Q Ch. 1.3 - Prob. 10Q Ch. 1.3 - Prob. 11Q Ch. 1.3 - Prob. 12Q Ch. 1.3 - Prob. 13Q Ch. 1.3 - Prob. 14Q Ch. 1.3 - Prob. 15Q Ch. 1.3 - Prob. 16Q Ch. 1.3 - Prob. 17Q Ch. 1.3 - Prob. 18Q Ch. 1.3 - Prob. 19Q Ch. 1.4 - Prob. 1E Ch. 1.4 - Prob. 2E Ch. 1.4 - Prob. 3E Ch. 1.4 - Prob. 4E Ch. 1.4 - Prob. 5E Ch. 1.4 - Prob. 6E Ch. 1.4 - Prob. 7E Ch. 1.4 - Prob. 8E Ch. 1.4 - Prob. 9E Ch. 1.4 - Prob. 10E Ch. 1.4 - Prob. 11E Ch. 1.4 - Prob. 12E Ch. 1.4 - Prob. 13E Ch. 1.4 - Prob. 14E Ch. 1.4 - Prob. 15E Ch. 1.4 - Prob. 16E Ch. 1.4 - Prob. 17E Ch. 1.4 - Prob. 18E Ch. 1.4 - Prob. 19E Ch. 1.4 - Prob. 20E Ch. 1.4 - 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Prob. 5E Ch. 1.5 - Prob. 6E Ch. 1.5 - Prob. 7E Ch. 1.5 - Prob. 8E Ch. 1.5 - Prob. 9E Ch. 1.5 - Prob. 10E Ch. 1.5 - Prob. 11E Ch. 1.5 - Prob. 12E Ch. 1.5 - Prob. 13E Ch. 1.5 - Prob. 14E Ch. 1.5 - Prob. 15E Ch. 1.5 - Prob. 16E Ch. 1.5 - Prob. 17E Ch. 1.5 - Prob. 18E Ch. 1.5 - Prob. 19E Ch. 1.5 - Prob. 20E Ch. 1.5 - Prob. 21E Ch. 1.5 - Prob. 22E Ch. 1.5 - Prob. 23E Ch. 1.5 - Prob. 24E Ch. 1.5 - Prob. 25E Ch. 1.5 - Prob. 26E Ch. 1.5 - Prob. 27E Ch. 1.5 - Prob. 28E Ch. 1.5 - Prob. 29E Ch. 1.5 - Prob. 30E Ch. 1.5 - Prob. 31E Ch. 1.5 - Prob. 32E Ch. 1.5 - Prob. 33E Ch. 1.5 - Prob. 34E Ch. 1.5 - Prob. 35E Ch. 1.5 - Prob. 36E Ch. 1.5 - Prob. 37E Ch. 1.5 - Prob. 38E Ch. 1.5 - Prob. 39E Ch. 1.5 - Prob. 40E Ch. 1.5 - Prob. 41E Ch. 1.5 - Prob. 42E Ch. 1.5 - Prob. 43E Ch. 1.5 - Prob. 44E Ch. 1.5 - Prob. 45E Ch. 1.5 - Prob. 46E Ch. 1.5 - Prob. 47E Ch. 1.5 - Prob. 48E Ch. 1.5 - Prob. 49E Ch. 1.5 - Prob. 50E Ch. 1.5 - Prob. 51E Ch. 1.5 - Prob. 52E Ch. 1.5 - Prob. 53E Ch. 1.5 - Prob. 54E Ch. 1 - Prob. 1CR Ch. 1 - Prob. 2CR Ch. 1 - Prob. 3CR Ch. 1 - Prob. 4CR Ch. 1 - Prob. 5CR Ch. 1 - Prob. 6CR Ch. 1 - Prob. 7CR Ch. 1 - Prob. 8CR Ch. 1 - Prob. 9CR Ch. 1 - Prob. 10CR Ch. 1 - Prob. 11CR Ch. 1 - Prob. 12CR Ch. 1 - Prob. 13CR Ch. 1 - Prob. 14CR Ch. 1 - Prob. 15CR Ch. 1 - Prob. 16CR Ch. 1 - Prob. 17CR Ch. 1 - Prob. 18CR Ch. 1 - Prob. 19CR Ch. 1 - Prob. 20CR Ch. 1 - Prob. 21CR Ch. 1 - Prob. 22CR Ch. 1 - Prob. 23CR Ch. 1 - Prob. 1CT Ch. 1 - Prob. 2CT Ch. 1 - Prob. 3CT Ch. 1 - Prob. 4CT Ch. 1 - Prob. 5CT Ch. 1 - Prob. 6CT Ch. 1 - Prob. 7CT Ch. 1 - Prob. 8CT Ch. 1 - Prob. 9CT Ch. 1 - Prob. 10CT Ch. 1 - Prob. 11CT Ch. 1 - Prob. 12CT Ch. 1 - Prob. 13CT Ch. 1 - Prob. 14CT Ch. 1 - Prob. 15CT Ch. 1 - Prob. 16CT Ch. 1 - Prob. 17CT Ch. 1 - Prob. 1CA Ch. 1 - Prob. 2CA Ch. 1 - Prob. 3CA Ch. 1 - Prob. 4CA Ch. 1 - Prob. 5CA Ch. 1 - Prob. 6CA Ch. 1 - Prob. 7CA Ch. 1 - Prob. 8CA Ch. 1 - Prob. 9CA

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