Sec 1 Basic Algebra: Worked Examples & Practice

Secondary 1 Mathematics · G3 focus · Study guide

Learn algebraic notation, substitution, like terms, brackets, fractional coefficients and common-factor factorisation through explained examples and an area explorer.

Concept → worked examples → practice16 min read · practice linked below

Algebra is a language for numbers

A letter can stand for an unknown number or a number that changes. If a notebook costs $p$ dollars, three notebooks cost $3p$ dollars. The letter represents a number, so the usual rules of arithmetic still apply.

Read each symbol carefully

$ab=a\times b$ , $a^2=a\times a$ , and $a^2b=a\times a\times b$ . Thus $3y=y+y+y$ , but $y^3=y\times y\times y$ . The expression $\frac{3+y}{5}$ means divide the whole sum by 5. A denominator must not be zero.

In $5x-3y+7$ , the terms are $5x$ , $-3y$ and $7$ . The coefficients of $x$ and $y$ are 5 and −3; 7 is the constant term. An expression has a value once its letters are given values. An equation also has an equals sign and states that two expressions have the same value.

Worked example 1: translate, then substitute

A notebook costs $p$ dollars. A pen costs $2 less than a notebook. Write an expression for two notebooks and three pens, then find the cost when $p=5$ .

A pen costs $p-2$ dollars. Three pens cost $3(p-2)$ , because all three cost the same amount.
Add the two notebook costs: $C=2p+3(p-2).$
Substitute 5 for every $p$ : $C=2(5)+3(5-2)=10+9=\boxed{19}.$

The total is $19. Directly, the notebooks cost $10 and the three $3 pens cost $9, which agrees. In this situation $p\geq2$ , so the pen price is not negative.

Collect like terms

Like terms have the same letters raised to the same powers. For example, $4x$ and $-7x$ are like terms. $x$ , $x^2$ and $y$ are different kinds of term. Add their coefficients only when the letter parts match: $4x-7x=-3x$ . Constants can be combined with other constants.

The letter part identifies the group; the signed coefficients tell you how many of that kind remain.

Worked example 2: keep each sign with its term

Simplify $7x-3y+4-2x+5y-9$ , then evaluate it when $x=-2$ , $y=3$ .

Group like terms without changing signs: $(7x-2x)+(-3y+5y)+(4-9).$
Combine each group: $\boxed{5x+2y-5}.$
Use brackets around the negative value: $5(-2)+2(3)-5=-10+6-5=\boxed{-9}.$

Check: substituting into the original gives $-14-9+4+4+15-9=-9$ , the same value. Writing $5x+2y$ as $7xy$ would change both the operation and the expression.

Expand by multiplying every term

The distributive law says $a(x+b)=ax+ab$ . One way to see it is to split a rectangle into two parts: the area of the whole equals the sum of the two smaller areas. This picture uses positive lengths; the same algebraic rule also works for zero and negative numbers.

Explore the distributive law

Three rows, each containing $x+2$ units, give $3(x+2)=3x+6$ . Change a factor or a width to see how both smaller areas change.

The two regions have the same height. Adding their areas is equivalent to multiplying the total width by that height. Use the labels for the dimensions; the picture is not to scale.

For a = 3, x = 4 and b = 2: 3 × (4 + 2) = 3 × 4 + 3 × 2 = 18.

Worked example 3: a negative multiplier

Simplify $-2(3x-5)+4x$ .

Multiply both terms in the bracket by −2: $(-2)(3x)+(-2)(-5)+4x=-6x+10+4x.$
The product of two negatives is positive, so the constant is $+10$ .
Combine the $x$ -terms: $\boxed{-2x+10}.$

Check at $x=2$ : the original is $-2(6-5)+8=6$ ; the simplified expression is $-4+10=6$ . This checks this value; the distributive-law steps justify the identity for every $x$ .

Worked example 4: fractional coefficients

Simplify $\frac{2x}{3}-\frac{3(x-5)}{2}$ .

Use the common denominator 6. Multiply each numerator and denominator by the same number: $\frac{4x}{6}-\frac{9(x-5)}{6}.$
Keep brackets around the entire numerator being subtracted: $\frac{4x-9(x-5)}{6}=\frac{4x-9x+45}{6}.$
Collect like terms: $\boxed{\frac{45-5x}{6}}.$

Check at $x=3$ : the original is $2-(-3)=5$ , and the final form is $(45-15)/6=5$ . An expression is not an equation: multiplying only the original expression by 6 would change its value. Here we used equivalent fractions instead.

Factorise by reversing expansion

To factorise, write a sum as a product. Find a factor that divides every term, place it outside a bracket, then divide each original term by that factor to find what stays inside. Taking the highest common factor leaves no further common factor inside.

Worked example 5: take out the highest common factor

Factorise $12x+18y$ completely.

The numerical HCF of 12 and 18 is 6. There is no common letter: the first term contains $x$ , while the second contains $y$ .
Divide each term by 6: $12x\div6=2x$ and $18y\div6=3y$ .
Write the product: $\boxed{6(2x+3y)}.$

Check by expanding: $6(2x+3y)=12x+18y$ . The answer $3(4x+6y)$ is equivalent but not fully factorised because another factor 2 can be removed from the bracket. If a letter is common too, include it: $8ab+12ac=4a(2b+3c)$ .

Common mistakes and the next step

Adding unlike terms: $3x+2$ stays as it is; it is not $5x$ .
Confusing powers and coefficients: $2x$ means $2\times x$ ; $x^2$ means $x\times x$ .
Missing a bracket term: $-3(x-2)=-3x+6$ , because −3 multiplies both terms.
Dropping a term when factorising: $5x+5=5(x+1)$ ; the constant contributes 1 inside the bracket.

Once the method makes sense, try the basic algebra practice bank. Use the linear equations guide when the task asks you to find an unknown.