E2.5 Equations

Learning objectives

E2.5.1Construct expressions, equations and formulas• Construct expressions, equations and formulas, including simultaneous equations.E2.5.2Solve linear equations• Solve linear equations in one unknown.E2.5.3Solve fractional equations• Solve fractional equations with numerical and linear algebraic denominators.E2.5.4Solve simultaneous linear equations• Solve simultaneous linear equations in two unknowns.E2.5.5Solve linear and non-linear simultaneous equations• Solve simultaneous equations involving one linear and one non-linear equation, with powers no higher than two.E2.5.6Solve quadratic equations• Solve quadratic equations by factorisation, completing the square and the quadratic formula; write completed-square forms and give solutions in surd form when required.E2.5.7Change the subject of formulas• Change the subject of formulas, including cases where the subject appears twice or is raised to a power or under a root.

Algebra models a situation by naming unknown quantities and translating relationships into expressions, equations or formulas. An expression has no equality sign; an equation states that two quantities are equal; a formula links several variables.

Words	Algebraic structure
5 more than $x$	$x+5$
5 less than $x$	$x-5$
5 less than twice $x$	$2x-5$
$y$ is 3 times $x$	$y=3x$
total is 40	add the parts and set the sum equal to 40

Define each unknown with its unit; build each quantity from that definition; use the relationship word to choose the operation; then check that both sides of an equation have the same meaning and units.

$\text{adult tickets}=x,\quad \text{child tickets}=x+6,\quad x+(x+6)=40$

When two unknowns are linked by two independent facts, construct two equations. For example, if $x+y=18$ and adult tickets cost $8 while child tickets cost$ 5, the revenue fact is $8x+5y=111$ .

Keep an equality only when the words assert equality. Do not write $2x+5=\,$ as an expression, and do not reverse phrases such as ‘5 less than $x$ ’.

Solving a linear equation means finding the value that makes both sides equal. Every valid step performs the same reversible operation on both sides, so the balance and the solution are preserved.

Expand brackets; clear numerical fractions if helpful; collect all terms containing the unknown on one side; collect constants on the other; divide by the coefficient; substitute the result into the original equation to check.

$3(2x-5)+4=2(x+7)\;\Longrightarrow\;6x-11=2x+14\;\Longrightarrow\;4x=25\;\Longrightarrow\;x=\frac{25}{4}$

Substitution gives $3(2\cdot25/4-5)+4=53/2$ and $2(25/4+7)=53/2$ , so the value satisfies both sides.

A term changes sign because the same term was added or subtracted on both sides—not because it ‘moves across’. Distribute a negative multiplier to every term inside its bracket.

A fractional equation can be converted into an ordinary linear or quadratic equation by multiplying every term by a common denominator. Values that make an original denominator zero are excluded from the start.

List excluded values; factor denominators if needed; choose a lowest common denominator; multiply every term on both sides by it; solve the resulting equation; reject any excluded or non-satisfying result.

$\frac1{x+1}+\frac9{x+9}=1,\quad x\ne-1,-9$

Multiplying by $(x+1)(x+9)$ gives $(x+9)+9(x+1)=(x+1)(x+9)$ . This simplifies to $x^2-9=0$ , so $x=3$ or $x=-3$ ; both are allowed and satisfy the original equation.

Cancelling a denominator is shorthand for multiplying every term by a non-zero expression. Never multiply only selected terms, and never accept a root that made an original denominator zero.

A solution to two simultaneous linear equations is one ordered pair that satisfies both equations. Elimination removes one unknown by combining aligned equations; substitution replaces one unknown with an equivalent expression.

Structure	Efficient method
matching or easily matched coefficients	elimination
one variable already isolated	substitution
neither is convenient	rearrange or scale first, then choose

$x+2y=13,\quad x+5y=22$

Subtracting the first equation from the second gives $3y=9$ , so $y=3$ . Substitution into $x+2y=13$ gives $x=7$ . The ordered pair is $(7,3)$ .

Check both originals: $7+2(3)=13$ and $7+5(3)=22$ . A pair is not a solution unless it passes both equations.

When subtracting equations, subtract every term, including signs and constants. Do not report separate unpaired values of $x$ and $y$ .

Substitution turns a linear–non-linear simultaneous system into one quadratic equation. Its two roots can produce two intersection points, and each root must be paired with its corresponding value of the other variable.

Rearrange the linear equation for one variable; substitute into the non-linear equation; expand and collect into a quadratic; solve it; substitute each root separately into the linear equation; verify each ordered pair in both originals.

$y=4-x,\quad x^2+2y^2=67$

Substitution gives $x^2+2(4-x)^2=67$ , so $3x^2-16x-35=0=(3x+5)(x-7)$ . Hence $x=7$ gives $y=-3$ , while $x=-5/3$ gives $y=17/3$ .

Do not mix the $y$ value from one root with the other $x$ root. A tangent may give one repeated solution; no real intersection gives no real ordered pair.

A quadratic equation can be solved by factorisation, completing the square or the quadratic formula. The equation must first be written as $ax^2+bx+c=0$ with $a
e0$.

Method	Best use	Result
factorisation	factors are visible or easy to find	exact roots
completing the square	vertex form or structural insight is useful	exact or surd roots
quadratic formula	works for every quadratic	exact or rounded roots

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

For $2x^2-x-6=0$ , $(2x+3)(x-2)=0$ , so $x=-3/2$ or $x=2$ . The zero-product rule applies only after one side equals zero.

For $x^2-6x+1=0$ , $(x-3)^2-8=0$ , so $x=3\pm\sqrt8=3\pm2\sqrt2$ . This is completed-square form and an exact surd solution.

The discriminant $b^2-4ac$ predicts two distinct real roots when positive, one repeated real root when zero, and no real roots when negative.

Keep the $\pm$ when taking a square root, place the whole numerator over $2a$ , and round only at the final step when a decimal accuracy is requested.

Changing the subject isolates a chosen variable using inverse operations while preserving equality. When the subject appears more than once, collect all its terms and factor it out before dividing.

Where the subject appears	Key move
once in a chain of operations	reverse the operations in a valid order
in a denominator	clear denominators first
in two or more terms	collect those terms, then factor
raised to a power	isolate the power, then take the appropriate root

$2mh=g(1-h)\;\Longrightarrow\;2mh=g-gh\;\Longrightarrow\;h(2m+g)=g\;\Longrightarrow\;h=\frac{g}{2m+g}$

$m=2p+\sqrt{\frac{x}{y}}\;\Longrightarrow\;m-2p=\sqrt{\frac{x}{y}}\;\Longrightarrow\;x=y(m-2p)^2$

From $A=\pi r^2$ , a radius is non-negative, so $r=\sqrt{A/\pi}$ . Without a contextual sign restriction, solving $x^2=k$ gives $x=\pm\sqrt{k}$ for $k\ge0$ .

Do not divide by a factor that could be zero without recording the restriction. Squaring can hide a sign condition, so check the rearranged formula against the original context.

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—E2.6 InequalitiesRead

E2.5 Equations

Learning objectives

Construct expressions, equations and formulas from information

Solve linear equations by preserving balance

Solve fractional equations and control excluded values

Solve simultaneous linear equations

Solve one linear and one non-linear equation simultaneously

Choose a method to solve quadratic equations

Change the subject of a formula

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