WASSCE Algebraic Expressions is one of the topics in WASSCE mathematics syllabus. The followings are WASSCE past questions and solutions on Algebraic Expressions. Candidates are advised to try each question by themselves before viewing solution. Frequent practice guarantees success in exams. VISIT MATHS LIBRARY FOR MORE TOPICS

WASSCE 2014 private candidates Factorize completely: \(m^2-2mn+n^2-9r^2.\)

Solution Solution:

\(m^2-2mn+n^2-9r^2.\)

\(=m^2-mn-mn+n^2-3^2r^2\)

\(=m(m-n)-n(m-n)-(3r)^2\)

\(=(m-n)(m-n)(3r)^2\)

\(=(m-n)^2-(3r)^2\)

\(=[(m-n)-3r][(m-n)+3r]\)

By: Abdul Karim

WASSCE 2012 private candidates. Factorize completely: \(9a^2-4(a-b)^2.\)

Solution Solution:

\(9a^2-4(a-b)^2.\)

\(=3^2a^2-2^2(a-b)^2\)

\(=(3a)^2-[2(a-b)]^2\)

\(=[3a-2(a-b)][3a+2(a-b)]\)

By: Abdul Karim

WASSCE 2011 private candidates. Simplify: \(\frac{m+1}{m-1}-\frac{m-1}{m+1}+\frac{4}{m^2-1}.\)

Solution Solution:

\(\frac{m+1}{m-1}-\frac{m-1}{m+1}+\frac{4}{m^2-1}\)

\(=\frac{m+1}{m-1}-\frac{m-1}{m+1}+\frac{4}{(m+1)(m-1)}\)

\(=\frac{(m+1)(m+1)-[(m-1)(m-1)]+4}{(m+1)(m-1)}\)

\(=\frac{m^2+2m+1-[m^2-2m+1]+4}{(m+1)(m-1)}\)

\(=\frac{m^2+2m+1-m^2+2m-1+4}{(m+1)(m-1)}\)

\(=\frac{m^2-m^2+2m+2m+1-1+4}{(m+1)(m-1)}\)

\(=\frac{4m+4}{(m+1)(m-1)}\)

\(=\frac{4(m+1)}{(m+1)(m-1)}\)

\(=\frac{4}{m-1}.\) Hint \((m+1)\) cancelled each other

By: Abdul Karim

WASSCE 2011 school candidates. Simplify: \(\frac{x+\frac{1}{x}+2}{x^2-1}.\)

Solution Solution:

\(\frac{x+\frac{1}{x}+2}{x^2-1}.\)

\(=\frac{\frac{x^2+1}{x}-2}{x^2-1}\)

\(=\frac{\frac{x^2+1+2x}{x}}{x^2-1}\)

\(=\frac{x^2+2x+1}{x(x^2-1)}\)

\(=\frac{x^2+x+x+1}{x[(x+1)(x-1)]}\)

\(=\frac{x(x+1)+1(x+1)}{x(x+1)(x-1)}\)

\(=\frac{(x+1)(x+1)}{x(x+1)(x-1)}\)

\(=\frac{x+1}{x(x-1)}.\) Hint \((x+1)\) cancelled each other

By: Abdul Karim

WASSCE 2010 private candidates. Factorize: \(x^2+4x+3+mx+3m.\)

Solution Solution:

\(x^2+4x+3+mx+3m\)

\(=x^2+3x+x+3+mx+3m\)

\(=x(x+3)+1(x+3)+m(x+3)\)

\(=(x+1+m)(x+3).\)

By: Abdul Karim

WASSCE 2009 school candidates. Simplify: \(\frac{x^2+x-6}{x^2-3x+2}\times\frac{x^2-x}{x^2-9}.\)

Solution Solution:

\(\frac{x^2+x-6}{x^2-3x+2}\times\frac{x^2-x}{x^2-9}\)

\(=\frac{x^2+3x-2x-6}{x^2-2x-x+2}\times\frac{x(x-1)}{x^2-3^2}\)

\(=\frac{x(x+3)-2(x+3)}{x(x-2)-1(x-2)}\times\frac{x(x-2)}{(x+3)(x-3)}\)

\(=\frac{(x+3)(x-2)}{(x-2)(x-1)}\times\frac{x(x-1)}{(x+3)(x-3)}\)

\(=\frac{x}{x-3}.\) Hint \((x+3)\), \((x-2)\) and \((x-1)\) have been cancelled

By: Abdul Karim

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