Algebra involves using symbols to represent unknown values and relationships. Solving an equation means finding every value of the unknown that makes the equation true. In examinations, show substitutions, factorisation and simplification clearly because marks are awarded for the method as well as the answer. Consistent accuracy is applied throughout the marking process.
A linear equation has an unknown raised only to the first power. Its general form is:
ax+b=0,\quad a\ne0
To solve a linear equation, collect terms containing the unknown on one side and constants on the other.
Question: Solve:
3x+7=25
Step 1: Identify the equation.
3x+7=25
Step 2: Subtract the constant from both sides.
3x=25-7
Step 3: Simplify.
3x=18
Step 4: Divide both sides by the coefficient.
x=\frac{18}{3}
Step 5: State the final answer.
Common Mistake: Moving a term across the equal sign without changing its sign correctly.
Practical Application: Linear equations can model costs, income, distance, mixtures and the number of items sold.
A quadratic equation has the general form:
ax^2+bx+c=0,\quad a\ne0
It can be solved by factorisation, completing the square or using the quadratic formula:
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
The expression under the square root is the discriminant:
\Delta=b^2-4ac
Question: Solve:
5x^2+2x=-9
Step 1: Write the equation in standard form.
5x^2+2x+9=0
Step 2: Identify the coefficients.
a=5,\quad b=2,\quad c=9
Step 3: Substitute into the quadratic formula.
x=\frac{-2\pm\sqrt{2^2-4(5)(9)}}{2(5)}
Step 4: Simplify.
x=\frac{-2\pm\sqrt{-176}}{10}
Step 5: Interpret the result.
\therefore\text{ there are no real roots}
Exam Tip: Always place the equation in standard form before identifying the values of the coefficients.
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Factorisation rewrites an algebraic expression as a product. The zero-product property states:
AB=0\Rightarrow A=0\text{ or }B=0
Question: Solve:
x^2+3x-10=0
Step 1: Find two numbers whose product is negative ten and whose sum is three.
5(-2)=-10
5+(-2)=3
Step 2: Factorise.
(x+5)(x-2)=0
Step 3: Equate each factor to zero.
x+5=0
x-2=0
Step 4: Solve each equation.
x=-5
x=2
Step 5: State both answers.
x=-5\text{ or }x=2
This factorisation and zero-product approach is consistent with the method used to solve equations in the marking guidance.
Inequalities requiring factorisation follow a similar process, but the final answer is an interval.
8x^2>2x
8x^2-2x>0
2x(4x-1)>0
The critical values are:
x=0\quad\text{and}\quad x=\frac14
A sign analysis gives:
x<0\quad\text{or}\quad x>\frac14
Common Mistake: Dividing an inequality by an expression containing the unknown. Its sign may be unknown. Factorise and use critical values instead.
An exponential equation contains the unknown in an exponent. Try to express both sides using the same base. If this is impossible, logarithms may be used.
Remember:
a^m=a^n\Rightarrow m=n,\quad a>0,\quad a\ne1
Question: Solve:
2(2^{2x})-9(2^x)+4=0
Step 1: Use a substitution.
k=2^x
Step 2: Rewrite the equation.
2k^2-9k+4=0
Step 3: Factorise.
(2k-1)(k-4)=0
Step 4: Solve for the substituted variable.
k=\frac12\quad\text{or}\quad k=4
Step 5: Replace the substituted variable.
2^x=\frac12\quad\text{or}\quad2^x=4
2^x=2^{-1}\quad\text{or}\quad2^x=2^2
Step 6: State the final answer.
x=-1\quad\text{or}\quad x=2
The substitution and factorisation method shown here follows the worked exponential equation solution.
Common Mistake: Forgetting to return to the original variable after solving the quadratic equation in the substituted variable.
A surd equation contains the unknown inside a square root. Isolate the surd before squaring. Squaring can introduce an extraneous solution, so every answer must be checked in the original equation.
Question: Solve:
\frac{1}{\sqrt{x}}+\frac{1}{x}=2
Step 1: Isolate the surd term.
\frac{1}{\sqrt{x}}=2-\frac{1}{x}
Step 2: Square both sides.
\frac{1}{x}=\left(2-\frac{1}{x}\right)^2
Step 3: Expand and simplify.
\frac{1}{x}=4-\frac{4}{x}+\frac{1}{x^2}
x=4x^2-4x+1
4x^2-5x+1=0
Step 4: Factorise.
(4x-1)(x-1)=0
Step 5: Find the possible solutions.
x=\frac14\quad\text{or}\quad x=1
Step 6: Check in the original equation.
For the first possible solution:
\frac{1}{\sqrt{\frac14}}+\frac{1}{\frac14}=2+4=6
Therefore, it is rejected.
For the second possible solution:
\frac{1}{\sqrt{1}}+\frac{1}{1}=1+1=2
Step 7: State the final answer.
x=1
The selection of the valid root after checking is essential in surd equations.
Simultaneous equations must be satisfied by the same values of the unknowns. Two linear equations may be solved by elimination or substitution.
Question: Solve simultaneously:
x+y=7
2x-y=5
Step 1: Add the equations to eliminate one unknown.
(x+y)+(2x-y)=7+5
Step 2: Simplify.
3x=12
Step 3: Solve for the first unknown.
x=4
Step 4: Substitute into the first equation.
4+y=7
y=3
Step 5: State the ordered solution.
(x;y)=(4;3)
Exam Tip: Choose elimination when coefficients are equal or easy to make equal. Choose substitution when one unknown is already isolated.
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These consist of one linear equation and one quadratic equation. Use the linear equation to express one unknown in terms of the other, substitute into the quadratic equation and solve.
Question: Solve simultaneously:
x=y+2
5xy=x^2+6
Step 1: Substitute the linear equation into the quadratic equation.
5(y+2)y=(y+2)^2+6
Step 2: Expand.
5y^2+10y=y^2+4y+4+6
Step 3: Write in standard form.
4y^2+6y-10=0
2y^2+3y-5=0
Step 4: Factorise.
(2y+5)(y-1)=0
Step 5: Solve for the second unknown.
y=-\frac52\quad\text{or}\quad y=1
Step 6: Calculate the corresponding values of the first unknown.
x=-\frac52+2=-\frac12
x=1+2=3
Step 7: State both ordered solutions.
(x;y)=\left(-\frac12;-\frac52\right)\quad\text{or}\quad(x;y)=(3;1)
This method and these ordered solutions correspond to the simultaneous-equation working provided.
Common Mistake: Matching an answer for one unknown with the incorrect corresponding answer for the other unknown.
An algebraic solution must show logical symbolic steps rather than relying only on a calculator or graph.
A reliable approach is:
For rational equations, exclude values that make a denominator zero:
\frac{x+1}{x-3}=2,\quad x\ne3
For quadratic inequalities, find the critical values and use a sign diagram or number line. Do not include critical values when the inequality is strict.
Exam Tip: A calculator answer without the required algebraic steps may not earn all the available marks.
Exact answers should normally be retained during calculations. Round only the final answer unless instructed otherwise.
Common instructions include:
Question: Solve correct to two decimal places:
5x^2+9x+2=0
Step 1: Identify the coefficients.
a=5,\quad b=9,\quad c=2
Step 2: Substitute into the quadratic formula.
x=\frac{-9\pm\sqrt{9^2-4(5)(2)}}{2(5)}
Step 3: Simplify.
x=\frac{-9\pm\sqrt{41}}{10}
Step 4: Use a calculator.
x=-0{,}259687\ldots\quad\text{or}\quad x=-1{,}540312\ldots
Step 5: Round to two decimal places.
x=-0{,}26\quad\text{or}\quad x=-1{,}54
These rounded values agree with the quadratic solutions shown in the marking guidance.
Remember: Do not round intermediate answers unnecessarily. Early rounding can cause the final answer to fall outside the accepted examination range.
