Answer ( x , y ) = ( 1 2 , 11 2 ) Explanation Given Information: y = − 3 x + 7 , y = 3 x + 4 Formula Used: Substitution Method Calculation: For equation. y = − 3 x + 7 Put the value of x in y, Hence, x 3 2 y -2 1 For equation. y = 3 x + 4 Put the value of x in y, Hence, x 0 1 y 4 7 Now, both lines meet at point Equation 1: y = − 3 x + 7 Equation 2: y = 3 x + 4 By substituting the value of y from equation 1 to equation 2. − 3 x + 7 = 3 x + 4 6 x = 3 x = 1 2 Now substituting the value of x in equation 1. y = − 3 x + 7 y = − 3 × 1 2 + 7 y = − 3 + 14 2 y = 11 2 Hence, both lines meet at point ( x , y ) = ( 1 2 , 11 2 )

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN: 9780547587776

Author: HOLT MCDOUGAL

Publisher: HOLT MCDOUGAL

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Question

Chapter 10.2, Problem 26E

To determine

To calculate linear system by graphing.

Expert Solution & Answer

Answer to Problem 26E

(x,y)=(12,112)

Explanation of Solution

Given Information: y=−3x+7 , y=3x+4

Formula Used:Substitution Method

Calculation:

For equation. y=−3x+7

Put the value of x in y, Hence,

x	3	2
y	-2	1

For equation. y=3x+4

Put the value of x in y, Hence,

x	0	1
y	4	7

Holt Mcdougal Larson Pre-algebra: Student Edition 2012, Chapter 10.2, Problem 26E

Now, both lines meet at point

Equation 1: y=−3x+7

Equation 2: y=3x+4

By substituting the value of y from equation 1 to equation 2.

−3x+7=3x+4

6x=3

x=12

Now substituting the value of x in equation 1.

y=−3x+7

y=−3×12+7

y=−3+142

y=112

Hence, both lines meet at point

(x,y)=(12,112)

Chapter 10.2, Problem 25E

Chapter 10.2, Problem 27E

Chapter 10 Solutions

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

Ch. 10.1 - Prob. 1C Ch. 10.1 - Prob. 2C Ch. 10.1 - Prob. 3C Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E

Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E Ch. 10.1 - Prob. 11E Ch. 10.1 - Prob. 12E Ch. 10.1 - Prob. 13E Ch. 10.1 - Prob. 14E Ch. 10.1 - Prob. 15E Ch. 10.1 - Prob. 16E Ch. 10.1 - Prob. 17E Ch. 10.1 - Prob. 18E Ch. 10.1 - Prob. 19E Ch. 10.1 - Prob. 20E Ch. 10.1 - Prob. 21E Ch. 10.1 - Prob. 22E Ch. 10.1 - Prob. 23E Ch. 10.1 - Prob. 24E Ch. 10.1 - Prob. 25E Ch. 10.1 - Prob. 26E Ch. 10.1 - Prob. 27E Ch. 10.1 - Prob. 28E Ch. 10.1 - Prob. 29E Ch. 10.1 - Prob. 30E Ch. 10.1 - Prob. 31E Ch. 10.1 - Prob. 32E Ch. 10.2 - Prob. 1C Ch. 10.2 - Prob. 2C Ch. 10.2 - Prob. 3C Ch. 10.2 - Prob. 1E Ch. 10.2 - Prob. 2E Ch. 10.2 - Prob. 3E Ch. 10.2 - Prob. 4E Ch. 10.2 - Prob. 5E Ch. 10.2 - Prob. 6E Ch. 10.2 - Prob. 7E Ch. 10.2 - Prob. 8E Ch. 10.2 - Prob. 9E Ch. 10.2 - Prob. 10E Ch. 10.2 - Prob. 11E Ch. 10.2 - Prob. 12E Ch. 10.2 - Prob. 13E Ch. 10.2 - Prob. 14E Ch. 10.2 - Prob. 15E Ch. 10.2 - Prob. 16E Ch. 10.2 - Prob. 17E Ch. 10.2 - Prob. 18E Ch. 10.2 - Prob. 19E Ch. 10.2 - Prob. 20E Ch. 10.2 - 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