Algebra Practice: Practical Guide & Workbook Exercises Fully Solved With Step By Step Explanation Sarah Nencini Author

Chapters of the book :Divisibility of Numbers, Integers, Factorization, Prime and Composite Numbers, Prime Factorization, Greatest Common Factor (GCF), Least Common Multiple (LCM), Fractions Decimals, Percents, Integer Exponents, Order Of Operation (... Read more

Chapters of the book :Divisibility of Numbers, Integers, Factorization, Prime and Composite Numbers, Prime Factorization, Greatest Common Factor (GCF), Least Common Multiple (LCM), Fractions Decimals, Percents, Integer Exponents, Order Of Operation (PEMDAS), Rational Exponents, Polynomials, Binomials, Radicals, Radical Expressions, Rational Expressions, Intro to Equations, Intro to Inequalities, Intro to Coordinate Plan, Linear Equations With One Variable, Absolute Value Equations, Rational Equation, Radical Equation, Quadratic Equations, Irrational Equations, Linear Equations & Functions, Systems of Equations, Systems of Linear and Quadratic Equations, Inequalities, Systems of Inequalities.Below some examples about how the book explains problem to solve and step by step solutionsProblem 1: 2(x+3)(x+1)=x+15Solution:th equation is not in the normal form,so we convert it to a normal form(2x+6)(x+1)=x+15 → 〖2x〗^2+2x+6x+6=x+15〖2x〗^2+2x+6x+6-x-15=0〖2x〗^2+7x-9=0a=2 b=7 c=-9∆=b^2-4ac=7^2-4(2)(-9)=49+72=121since ∆ 0 (positive) we will have two real different solutions:x_1,x_2=(-b±√(b^2-4ac))/2a=(-b±√∆)/2a=(-(7)±√121)/2(2) =(-7±√121)/4=(-7±11)/4x_1=(-7+11)/4=4/4=1 → x_1=1x_2=(-7-11)/4=-18/4=-9/2 → x_2=-9/2Problem 2: Solve the Radical Expression: 2√2+√18-√8+√50SolutionIn this case there are no perfect square of the radicands,so we have to apply the prime factorization to each radicand and then move the factors with exponents 2 out of the radical18=2∙3^2→√18=√(2∙3^2 )=3√28= 2^3→√8=√(2^2∙2)=2√250=2∙5^2→√50=√(2∙5^2 )=5√2we can so rewrite the expression: 2√2+3√2-2√2+5√2=(2+3-2+5) √2=8√2Problem 3: Solve the equation x^2+13x=0Solution:Factor → x(x+13)=0Apply Zero property x_1=0x_2+13=0 → x_2=-13Problem 4: Solve the rational equation 1/(x-2)=5/xSolution:1/(x-2)=5/x LCD= x(x-2) → Restrictions: x≠0 and x-2 ≠0 → x≠2The denominator can be eliminated because we can multiply both sides by x(x-2) x(x-2)∙1/(x-2)=x(x-2)∙ 5/xx=(x-2)∙5 x=5x-10 x-5x=-10 -4x=-10 we can change both signs4x=10 x=10/4=5/2Solution: x=5/2The solution is determinated since 5/2 satisfies the restrictionsProblem 5: Solve the equation (4-2x)/3=3/4-5x/6Solution Find the LCD,for 3,4 and 6,then execute the operation: LCD(3,4,6)=12(4(4-2x))/12=(3∙3-2(5x))/12now multiply both memebers by 12 (to delete the denominators) and simplify 16-8x=9-10x-8x+10x=-16+92x=-7x=-7/2 Less