Solved Exercises Problems in System of Linear Equations in One Variable-Algebra

COLLEGE ALGEBRA ǀ MATH 114 ǀ NOTRE DAME UNIVERSITY
PROBLEMS WITH SOLUTIONS TO COLLEGE ALGEBRA TEXTBOOK BY PAUL R. RIDER
SYSTEMS OF LINEAR EQUATIONS in ONE VARIABLE - SOLVED EXERCISES PROBLEMS IN
COLLEGE ALGEBRA
EXERCISES II. A
Solve for x if possible:
1.) 7x + 42 = 0
SOLUTION: STEPS:
7x + 42 = 0
7x = -42 transpose constant term and change its sign
1
( 7x = -42 ) ( ) multiply both side by constant’s reciprocal of x
7

x=-6 and evaluate for the value of x as a Solution if any
Since x has a value of -6, then it is the solution that we will see on the graph using Geogebra App
2.) 2x + 19 = 35
SOLUTION: STEPS:
2x + 19 = 35
2x = 35 – 19 transpose constant term and change its sign
2x = 16 perform arithmetic operations
1
( 2x = 16 ) (2) multiply both side by constant’s reciprocal of x

x=8 and evaluate for the value of x as a Solution if any
Since x has a value of 8, then it is the solution that we will see on the graph using Geogebra App
3.) 5x – 8 = 2x + 7
SOLUTION: STEPS:
5x – 8 = 2x + 7
5x – 2x = 7 + 8 transpose constant term and change its sign
3x = 15 perform arithmetic operations
1
( 3x = 15) (3) multiply both side by constant’s reciprocal of x

x=5 and evaluate for the value of x as a Solution if any
Since x has a value of 5, then it is the solution that we will see on the graph using Geogebra App

, 4.) 6x + 13 = 11x – 10
SOLUTION: STEPS:
6x + 13 = 11x – 10
6x – 11x = – 10 – 13 transpose constant term and collect like terms
-5x = – 23 perform arithmetic operations for like terms
1
( -5x = -23) ( ) multiply both side by constant’s reciprocal of x
−5

𝟐𝟑 𝟑
x= or 𝟒 𝟓 and evaluate for the value of x as a Solution if any
𝟓

Since x has a value of 23/5 , then it is the solution that we will see on the graph using Geogebra App
5.) 3x + 2 - 4(x-3) = 2(5x – 4)
SOLUTION: STEPS:
3x + 2 - 4(x-3) = 2(5x – 4)
3x + 2 - 4x+12 = 10x – 8 apply distribution property
3x – 4x – 10x = -8 – 12 - 2 transpose constant term and collect like terms
-11x = -22
1
( -11x = -22) (−11) multiply both side by constant’s reciprocal of x

x=2 and evaluate for the value of x as a Solution if any
Since x has a value of 2 , then it is the solution that we will see on the graph using Geogebra App
6.) x2 + 4x + 5 = (x+2)(x+7)
SOLUTION: STEPS:
x2 + 4x + 5 = (x+2)(x+7)
x2 + 4x + 5 = 7x + 2x + 14 simplify the expression
x2 - x2 + 4x - 7x - 2x = 14 – 5 transpose constant term and collect like terms
-5x = 9
1
( -5x = 9) ( ) multiply both side by constant’s reciprocal of x
−5

𝟗 𝟒
x = − 𝟓 or −𝟏 𝟓 and evaluate for the value of x as a Solution if any

Since x has a value of -9/5 , then it is the solution that we will see on the graph using Geogebra App